Note that the triple $(k,1,1)$ works where $k$ is any positive integer and $m=1,n=1$. We prove it is the only one. Rearranging gives $$m^n-1=k^m(k^{n-m}-1)$$. We can use a convexity argument here. The right side grows much faster than the left, so to minimize we need solutions close to $0$. We see that only $1,1$ for $m,n$ work since it effectively makes both sides $0$.

S.Das93 wrote: Note that the triple $(k,1,1)$ works where $k$ is any positive integer and $m=1,n=1$. We prove it is the only one. Rearranging gives $$m^n-1=k^m(k^{n-m}-1)$$. We can use a convexity argument here. The right side grows much faster than the left, so to minimize we need solutions close to $0$. We see that only $1,1$ for $m,n$ work since it effectively makes both sides $0$. doesn't the triple $(1,1,n)$ also work?

I claim the only solutions are $(m,n,k)=(1,1,k)$ and $(1,n,1)$, where $k,m$ are arbitrary. First, for $k=1$, we get $m^n=1$, so $(1,n,1)$ works for any $n$. Assume $k>1$. Furthermore, if $m=n$, then $m^n=1$ forcing $m=n=1$, so $(1,1,k)$ is also a valid family. We now show there are no other solutions. Note that $k^n = m^n+k^m-1>m^n$ (as $k^m>1$), forcing $k\ge m+1$. Now, \[ m^n-1 = k^m\left(k^{n-m}-1\right)\ge (m+1)^m \left((m+1)^{n-m}-1\right)>(m+1)^m m^{n-m}>m^n. \]This is a clear contradiction.

If $m=1$ then $k+1=k^n+1\iff n=1\vee k=1$, from now on $m>1$ and we will show that then there are no solutions. We have $k^n-k^m =m^n-1 >0\implies n>m$. Also $m^n-1=k^n-k^m\le k^n-1 \implies m\le k\implies k\ge 2\implies m<k$. Also $m^n>m^n-1=k^n-k^m \ge k^n-k^{n-1}=k^n(1-{1\over k})\ge {{k^n}\over 2}\implies (k/m)^n\le 2\implies (1+1/m)^n \le 2 \implies (1+1/m)^m \le 2 \implies m=1$. Indeed we have $(1+1/m)^m\ge 1+m/m=2$ by the binomial expansion and more terms are not allowed.