Assume that $x^3+1$ factored as a product of two nontrivial functors $f(x)$ and $g(x)$. Then if we took $X=[n]:=\{1,2,\cdots,n\},$ we would need the $S_n$-set $[n]^3 \sqcup \{*\}$ to be a product of two $S_n$-sets $f([n])$ and $g([n])$. Looking at $S_n$-fixed points, we see that for $n\geq 3$ $f([n])$ (resp. $g([n])$) must have a unique $S_n$-fixed point, which we will denote $*_f$ (resp. $*_g$).
If $f([n])\neq\{*_f\},$ then $f([n])$ must have size at least $n$ (by the classification of small $S_n$-sets used below), and so $g([n])$ must have size at most $n^2$. Take $n$ sufficiently large. Then, there is a classification of $S_n$-sets with at most $n^2$ elements (see https://math.stackexchange.com/questions/3536888/large-subgroups-of-symmetric-group). In particular, each $S_n$-orbit with at most $n^2$ elements has an $A_{n-2}$ fixed point, and $[n]^3+1$ has $9$ $A_{n-2}$ fixed points, so $f([n])$ and $g([n])$ have at most $9$ $S_n$-orbits each. It follows that, for a fixed $n$, the size of $f([n])$ must be the value at $n$ of one of a finite number of polynomials. Therefore, for a fixed (still sufficiently large $n$), $f([n])$ and $g([n])$ either have sizes $1$ and $n^3+1$ or $n+1$ and $n^2-n+1$.
Assume that for all sufficiently large $n$, $f([n])$ and $g([n])$ are nontrivial. Fix a value of $n$ and assume WLOG that $f([n])$ has $n+1$ elements. By the classification above, it must be isomorphic as a $S_n$-set to the union of $*_f$ and $S_n/S_{n-1}.$ Similarly, using that $g([n])$ has at most $9$ $S_n$-orbits, the classification tells us that $g([n])$ must be a union of $*_g$ and orbits of the form $S_n/G$, where $G$ is $S_{n-2}$, $S_{n-2}\times S_2$,$A_{n-2}\times S_2$, or the index $2$ subgroup $H:=\{\sigma,\operatorname{sign}(\sigma)\}\subseteq S_{n-2}\times S_2.$ Taking invariants with respect to $A_{n-2}\times S_2$ rules out the middle two cases and taking invariants with respect to $H$ rules out the last case, so we must have $g([n])\cong{*_g}\cup S_n/S_{n-2}$ as a $S_n$-set.
Now consider the map $[n]\rightarrow[n-1]$ which sends $i$ to $i$ for $i\leq n-1$ and sends $n$ to $n-1.$ If $f([n-1])$ had size $(n-1)^2-(n-1)+1$, then the map $f([n])\rightarrow f([n-1])$ could not be surjective, which contradicts the observation that $[n]^3+1\rightarrow[n-1]^3+1$ is surjective. So we can assume $f([n-1])$ has size $n$ (and so the previous paragraph applies to $f([n-1])$ and $g([n-1])$). There are three $S_{n-2}$-invariant points of $g([n])$, and the map $g([n])\rightarrow g([n-1])$ must send them all to $*_g$ (the only $S_{n-2}$-invariant point of $g([n-1])$). But that would imply that the fiber of $*$ along $[n]^3+1\rightarrow[n-1]^3+1$ has size at least $3$, which is not true (it consists only of $*$).
We have shown that we can find arbitrary large $n$ such that $f([n])$ or $g([n])$ is trivial. WLOG, assume that there exists arbitrary large $n$ with $f([n])$ trivial. Then for any $[m]$, if we choose an injection $[m]\rightarrow[n]$ with $f([n])$ trivial, the map $f([m])\rightarrow f([n])$ must be an injection, so $f([m])$ must be trivial, as desired.
