Consider modulo $9$, we have $3^m \equiv 1,3$ or $0$, and $s(n) + 1182 \equiv n + 3$, so $n^2 \equiv 3,4$ or $6$, but the only possible residues of $n^2 \pmod 9$ are $0,1,4,7$, so $m = 0$ is the only option. And so $n(n+1) = s(n) + 1183$, so $n(n+1) > 1183$, and so $n \geq 34$. We also know that $s(n) \leq n$, so $n^2 \leq 1183$, and so $n \leq 34$. So the only option is $34$, which works. That means the only solution is $(m,n) = (0,34)$