To fully understand logarithms of nonpositive numbers, you you need to be familiar with the exponential forms of complex numbers. Let $z = a+ib$ with $a,b \in \mathbf R$. Then, $e^{a+ib} = e^a e^{ib}$. $e^a$ is a real, nonnegative number, while $e^{ib}$ is equal to $\cos b + i\sin b$. In other words, to find $e^z$ on the complex plane, you take a point with abscissa $e^z$ on the real axis, and you rotate your plane by $b$ radians. You can get any point on the plane, except $0$ (do you see why?). For instance, $i = e^{0 + \frac \pi 2 i}$, because $i$ has a distance $1 = e^0$ of the origin, and is located at an angle $\pi/2$ radians from the real axis. The same way, $-1 = e^{\pi i}$.
This said, the logarithm of any complex number $Z$ (including real, negative numbers) should be a number $z$ such that $e^z = Z$. A first issue is that there is no uniqueness of such number. Indeed, rotating by $2\pi$ radians does not change anything. Hence, if $e^{a + bi} = Z$, then $e^{a + bi + 2\pi i} = Z$. So do $e^{a + bi + 2\ell\pi i}$, for every integer $\ell$: adding $2\ell \pi i$ is equivalent to doing $\ell$ rounds, i.e. changing nothing $\ell$ times. For instance, $-1 = e^{\pi i} = e^{-\pi i} = e^{3\pi i}$. Should the logarithm of $-1$ be equal to $\pi i$, $-\pi i$ or $3\pi i$?
There is no good answer. In fact, you can choose which answer you want. In other words, you can choose a logarithm function $\mathbf C^* \longrightarrow \mathbf C$, which gives you a possible logarithm for every nonzero number of $\mathbf C$. But you have to choose in which interval of length $2\pi$ you want to choose the ordinates. A very common choice is to take the ordinates in the interval $(-\pi, \pi]$. With this convention, you can write that $\ln(-1) = 0 + \pi i$, $\ln(-2) = \ln(2) + \pi i$, etc.
But this is a pure convention. Furthermore, this leads to some continuity issues: indeed, $\ln(-1) = \pi i$, but if $\varepsilon > 0$ is very close to zero, then $\ln(-1 - \varepsilon i)$ is very close to $-\pi i$. In other words, the logarithm function “jumps” every time you cross the half-line of negative real numbers.
Now, let’s focus on your proposal. Say $k$ is such that $2^k = -2$. This can be rewritten: $(e^{\ln 2})^k = e^{i \pi}e^{\ln 2}$. Hence, $e^{k \ln 2} = e^{\ln 2 + i\pi}$.
But from what I previously said about $e^{a + bi}$ being a point at distance $e^a$ rotated by $b$ radians, we can infer the following equivalence: for every complex numbers $w$ and $z$, we have
$$e^w = e^z \iff \exists \ell \in \mathbf Z : z = w+ 2\ell \pi i$$
So with the $k$ you defined, one must have:
$$k \ln 2 = \ln 2 + \pi i + 2\ell \pi i, \quad \ell \in \mathbf Z$$
$$k = 1 + \frac{\pi + 2\ell \pi}{\ln 2} i, \quad \ell \in \mathbf Z$$
Now, assume we also have: $3^k = -3$. The same computations ensure that:
$$k \ln 3 = \ln 3 + \pi i + 2\ell' \pi i, \quad \ell' \in \mathbf Z$$
$$k = 1 + \frac{\pi + 2\ell' \pi}{\ln 3} i, \quad \ell' \in \mathbf Z$$
So there would exist integers $\ell$ and $\ell'$ such that
$$k = 1 + \frac{\pi + 2\ell \pi}{\ln 2} i = 1 + \frac{\pi + 2\ell' \pi}{\ln 3} i$$
In particular, taking the imaginary part, this leads to:
$$\frac{\pi + 2\ell \pi}{\ln 2} = \frac{\pi + 2\ell' \pi}{\ln 3}$$
Divide by $\pi$:
$$\frac{1 + 2\ell}{\ln 2} = \frac{1 + 2\ell'}{\ln 3}$$
This said:
$$(1+2\ell) \ln 3 = (1+2\ell') \ln 2$$
Now taking the exponential:
$$3^{1 + 2\ell} = 2^{1+2\ell'}$$
This is clearly impossible, since there exists no number which is a power of $2$ and $3$ at the same time (except $1$).
Maybe there exists easier ways to generate contradictions, but this one is enough to prove that there could not exist a number $k$ such that $x^k = -x$ for all real $x$.
