The Fermat curves (and FLT) are used in the quoted paper to prove the following result:

Theorem 3.1: There exists a (recursive) infinite sequence $\mathcal{C} = \{C_i\}$ of curves $C_i/\mathbb{Q}$ such that:
(i) for all $_i$, one can effectively determine the set $C_i(\mathbb{Q})$ and
(ii) $\mathcal{C}$ is non-covering: for all $i \neq j$, there does not exist a finite $k$-morphism $C_i \rightarrow C_j$.

As the authors remark, any family of nonisomorphic curves of common genus $g \geq 2$ is noncovering, so it would suffice to exhibit such an infinite family of curves such that each one provably has no $\mathbb{Q}$-rational points.

This is easy, and we do so below for each $g \geq 2$:

Start with the (smooth projective model of the) curve given by the following equation.

$C: y^2 = - (x^{2g+2} + 1)$.

This is a hyperelliptic curve of genus $g$ which is visibly without $\mathbb{R}$-points, so certainly $C(\mathbb{Q}) = \varnothing$.

Now, among many other constructions, one can perturb the coefficients of $C$ slightly to get infinitely other genus $g$ curves without $\mathbb{Q}$-rational points. Indeed, write $P(x) = -x^{2g+2} - 1$ and for $a = (a_1,\ldots,a_{2g+2}) \in \mathbb{Q}^{2g+2}$, consider the polynomial

$P_a(x) = P(x) + \sum_{i=0}^{2g+2} a_i x^i$.

There exists an effectively computable $\epsilon > 0$ such that if for all $i$, $|a_i|_{\infty} < \epsilon$, then $P_a(x)$ is negative for all real $x$, so the curve $C_a: y^2 = P_a(x)$ still has no $\mathbb{R}$-points and hence no $\mathbb{Q}$-points. Here $|x|_{\infty}$ is the standard Archimedean norm.

It is no problem to find infinitely many choices of $a$ leading to nonisomorphic curves $C_a$. For instance, one can find choices of $a$ which give rise to curves with pairwise distinct moduli. Alternately, by a standard weak approximation / Krasner's Lemma argument one can choose the coefficients so as to make the curve $C_a$ have / not have $\mathbb{Q}_p$-points for each prime $p$ in any finite set $S$ of rational primes. More details for this can be supplied on request.

There are many other possible constructions. For instance, in this paper I produced for each number field $K$ and each $g \geq 2$ a bielliptic curve $C_{/\mathbb{Q}}$ of genus $g$ which violates the Hasse Principle. In fact by varying the elliptic curve that $C$ maps $2:1$ onto, one gets infinitely many such examples. Note that here the curves have points over every completion but still no $\mathbb{Q}$-points. In general, producing curves which do not have $\mathbb{Q}$-points "for local reasons" is much easier.

In my opinion emphasizing the connection to Fermat's Last Theorem here is a bit misleading.

The Fermat curves (and FLT) are used in the quoted paper to prove the following result:

Theorem 3.1: There exists a (recursive) infinite sequence $\mathcal{C} = \{C_i\}$ of curves $C_i/\mathbb{Q}$ such that:
(i) for all $_i$, one can effectively determine the set $C_i(\mathbb{Q})$ and
(ii) $\mathcal{C}$ is non-covering: for all $i \neq j$, there does not exist a finite $k$-morphism $C_i \rightarrow C_j$.

As the authors remark, any family of nonisomorphic curves of common genus $g \geq 2$ is noncovering, so it would suffice to exhibit such an infinite family of curves such that each one provably has no $\mathbb{Q}$-rational points.

This is easy, and we do so below for each $g \geq 2$:

Start with the (smooth projective model of the) curve given by the following equation.

$C: y^2 = - (x^{2g+2} + 1)$.

This is a hyperelliptic curve of genus $g$ which is visibly without $\mathbb{R}$-points, so certainly $C(\mathbb{Q}) = \varnothing$.

Now, among many other constructions, one can perturb the coefficients of $C$ slightly to get infinitely other genus $g$ curves without $\mathbb{Q}$-rational points. Indeed, write $P(x) = -x^{2g+2} - 1$ and for $a = (a_1,\ldots,a_{2g+2}) \in \mathbb{Q}^{2g+2}$, consider the polynomial

$P_a(x) = P(x) + \sum_{i=0}^{2g+2} a_i x^i$.

There exists an effectively computable $\epsilon > 0$ such that if for all $i$, $|a_i|_{\infty} < \epsilon$, then $P_a(x)$ is negative for all real $x$, so the curve $C_a: y^2 = P_a(x)$ still has no $\mathbb{R}$-points and hence no $\mathbb{Q}$-points. Here $|x|_{\infty}$ is the standard Archimedean norm.

It is no problem to find infinitely many choices of $a$ leading to nonisomorphic curves $C_a$. For instance, one can find choices of $a$ which give rise to curves with pairwise distinct moduli. Alternately, by a standard weak approximation / Krasner's Lemma argument one can choose the coefficients so as to make the curve $C_a$ have / not have $\mathbb{Q}_p$-points for each prime $p$ in any finite set $S$ of rational primes. More details for this can be supplied on request.

There are many other possible constructions. For instance, in this paper I produced for each number field $K$ and each $g \geq 2$ a bielliptic curve $C_{/\mathbb{Q}}$ of genus $g$ which violates the Hasse Principle. In fact by varying the elliptic curve that $C$ maps $2:1$ onto, one gets infinitely many such examples. Note that here the curves have points over every completion but still no $\mathbb{Q}$-points. In general, producing curves which do not have $\mathbb{Q}$-points "for local reasons" is much easier.

In my opinion emphasizing the connection to Fermat's Last Theorem here is a bit misleading.

The Fermat curves (and FLT) are used in the quoted paper to prove the following result:

Theorem 3.1: There exists a (recursive) infinite sequence $\mathcal{C} = \{C_i\}$ of curves $C_i/\mathbb{Q}$ such that:
(i) for all $_i$, one can effectively determine the set $C_i(\mathbb{Q})$ and
(ii) $\mathcal{C}$ is non-covering: for all $i \neq j$, there does not exist a finite $k$-morphism $C_i \rightarrow C_j$.

As the authors remark, any family of nonisomorphic curves of common genus $g \geq 2$ is noncovering, so it would suffice to exhibit such an infinite family of curves such that each one provably has no $\mathbb{Q}$-rational points.

This is easy, and we do so below for each $g \geq 2$:

Start with the (smooth projective model of the) curve given by the following equation.

$C: y^2 = - (x^{2g+2} + 1)$.

This is a hyperelliptic curve of genus $g$ which is visibly without $\mathbb{R}$-points, so certainly $C(\mathbb{Q}) = \varnothing$.

Now, among many other constructions, one can perturb the coefficients of $C$ slightly to get infinitely other genus $g$ curves without $\mathbb{Q}$-rational points. Indeed, write $P(x) = -x^{2g+2} - 1$ and for $a = (a_1,\ldots,a_{2g+2}) \in \mathbb{Q}^{2g+2}$, consider the polynomial

$P_a(x) = P(x) + \sum_{i=0}^{2g+2} a_i x^i$.

There exist an effectively computable $\epsilon > 0$ such that if for all $i$, $|a_i|_{\infty} < \epsilon$, then $P_a(x)$ is negative for all real $x$, so the curve $C_a: y^2 = P_a(x)$ still has no $\mathbb{R}$-points and hence no $\mathbb{Q}$-points. Here $|x|_{\infty}$ is the standard Archimedean norm.

It is no problem to find infinitely many choices of $a$ leading to nonisomorphic curves $C_a$. For instance, one can find choices of $a$ which give rise to curves with pairwise distinct moduli. Alternately, by a standard weak approximation / Krasner's Lemma argument one can choose the coefficients so as to make the curve $C_a$ have / not have $\mathbb{Q}_p$-points for each prime $p$ in any finite set $S$ of rational primes. More details for this can be supplied on request.

There are many other possible constructions. For instance, in this paper I produced for each number field $K$ and each $g \geq 2$ a bielliptic curve $C_{/\mathbb{Q}}$ of genus $g$ which violates the Hasse Principle. In fact by varying the elliptic curve that $C$ maps $2:1$ onto, one gets infinitely many such examples. Note that here the curves have points over every completion but still no $\mathbb{Q}$-points. In general, producing curves which do not have $\mathbb{Q}$-points for local reasons is much easier to do.

In my opinion emphasizing the connection to Fermat's Last Theorem here is a bit misleading.

The Fermat curves (and FLT) are used in the quoted paper to prove the following result:

Theorem 3.1: There exists a (recursive) infinite sequence $\mathcal{C} = \{C_i\}$ of curves $C_i/\mathbb{Q}$ such that:
(i) for all $_i$, one can effectively determine the set $C_i(\mathbb{Q})$ and
(ii) $\mathcal{C}$ is non-covering: for all $i \neq j$, there does not exist a finite $k$-morphism $C_i \rightarrow C_j$.

As the authors remark, any family of nonisomorphic curves of common genus $g \geq 2$ is noncovering, so it would suffice to exhibit such an infinite family of curves such that each one provably has no $\mathbb{Q}$-rational points.

This is easy, and we do so below for each $g \geq 2$:

Start with the (smooth projective model of the) curve given by the following equation.

$C: y^2 = - (x^{2g+2} + 1)$.

This is a hyperelliptic curve of genus $g$ which is visibly without $\mathbb{R}$-points, so certainly $C(\mathbb{Q}) = \varnothing$.

Now, among many other constructions, one can perturb the coefficients of $C$ slightly to get infinitely other genus $g$ curves without $\mathbb{Q}$-rational points. Indeed, write $P(x) = -x^{2g+2} - 1$ and for $a = (a_1,\ldots,a_{2g+2}) \in \mathbb{Q}^{2g+2}$, consider the polynomial

$P_a(x) = P(x) + \sum_{i=0}^{2g+2} a_i x^i$.

There exists an effectively computable $\epsilon > 0$ such that if for all $i$, $|a_i|_{\infty} < \epsilon$, then $P_a(x)$ is negative for all real $x$, so the curve $C_a: y^2 = P_a(x)$ still has no $\mathbb{R}$-points and hence no $\mathbb{Q}$-points. Here $|x|_{\infty}$ is the standard Archimedean norm.

It is no problem to find infinitely many choices of $a$ leading to nonisomorphic curves $C_a$. For instance, one can find choices of $a$ which give rise to curves with pairwise distinct moduli. Alternately, by a standard weak approximation / Krasner's Lemma argument one can choose the coefficients so as to make the curve $C_a$ have / not have $\mathbb{Q}_p$-points for each prime $p$ in any finite set $S$ of rational primes. More details for this can be supplied on request.

There are many other possible constructions. For instance, in this paper I produced for each number field $K$ and each $g \geq 2$ a bielliptic curve $C_{/\mathbb{Q}}$ of genus $g$ which violates the Hasse Principle. In fact by varying the elliptic curve that $C$ maps $2:1$ onto, one gets infinitely many such examples. Note that here the curves have points over every completion but still no $\mathbb{Q}$-points. In general, producing curves which do not have $\mathbb{Q}$-points "for local reasons" is much easier.

In my opinion emphasizing the connection to Fermat's Last Theorem here is a bit misleading.