The generating function $g(x):=(1−k^2 x)^{-1/k}$ satisfies, besides $g(0)=1,$
$ g^k= 1 + k^2 x\ g^k $,
whence we may express $c(k,n)$ as a sum of products of $c(k,j)$, with $j < n$, showing inductively that they are all integers, and in fact, multiples of $k$ for $n>0$.
Indeed, hiding the variable $k$ in $c(k,n)=c(n)$, one has
$c(0)=1$, $c(1)=k$
and in general for n>1,
$c(n)= k\sum_\mu c(\mu_1)c(\mu_2)..c(\mu_k) - \frac{1}{k}\sum_\nu c(\nu_1)c(\nu_2)..c(\nu_k)$
the first sum being extended over all multi-indices $\mu\in \mathbb{N}^k$ with weight $|\mu|:=\mu_1+\mu_2\dots +\mu_k=n-1$, while the second over all $\nu\in \mathbb{N}^k$ with $|\nu|=n$ and $\nu_j< n$ for $j=1\dots k$. It follows that if $c(j)$ are multiple of $k$ for $1 < j < n$, so is $c(n)$, and by induction this proves the claim. (The factor 1/k doesn't bother, because each term in the second sum contains at least 2 factors $c(j)$ with $0 < j < n$).