For $n=1$, any value of $k$ works. Suppose $n \geqslant 2$. Let $q$ be the smallest prime factor of $n$. We know that $q$ divides $k^{n}-1$, so $k^{n} \equiv 1(\bmod q)$. Therefore, $k$ is coprime with $q$, and its order (denoted $w$) divides $n$. By Fermat's theorem, the order divides $q-1$. Since $w$ divides $n$ and $q-1$, its prime factors are smaller than those of $q-1$ and thus strictly less than $q$. Moreover, they are also prime factors of $n$ and therefore greater than or equal to $q$. In particular, $w$ has no prime factors, so $w=1$. By the definition of the order, we deduce that $q$ divides $k^{1}-1=k-1$ and $q$ divides $n$, which contradicts the fact that $k-1$ and $n$ are coprime.

In particular, all the solution pairs are $(1, n)$ for $n$ a strictly positive integer.
Find all pairs $(k, n)$ of strictly positive integers such that $n$ and $k-1$ are coprime, and $n$ divides $k^{n}-1$.