Find all triplets of positive integers $(m, n, k)$ such that $k^{m} \mid m^{n}-1$ and $k^{n} \mid n^{m}-1$.
If $m=1$ then for $k^{n} \mid n-1$ it is necessary that $k=1$ or $n=1$.

If $m, n>1$, we can assume that $m \geq n$. Let $p$ be a prime divisor of $k$. Then $m$ and $n$ are both coprime with $p$.

If $p=2$, then $m$ and $n$ are odd. But then $1+m+\cdots+m^{n-1}$ is odd, so $2^{m} \mid m-1$. This is impossible because $2^{m}>m-1$. Therefore, $k$ is not divisible by 2.

Suppose that $p$ is odd. First, $p^{m} \mid m^{n}-1 \leq m^{m}-1$ so $pm^{p}$ (which is proven by studying the function $x \mapsto x^{1 / x}$), we have a contradiction. Therefore, $k$ has no odd prime divisor.

The solutions are thus $(m, n, 1)$ with $m, n$ arbitrary and $(1,1, k)$ with $k$ arbitrary.