Nice one. Note that $m\ge 4$. We have $(-1)^n \equiv 1(5)$ hence $n$ is even and $(3^\nu+1)(3^\nu-1)=5\cdot 2^m$. Then $3^\nu\pm 1$ are both even and $2||3^\nu+1$ or $2||3^\nu-1$, since the difference is 2. Hence $3^\nu+1\le 10 \vee 3^\nu-1\le 10 \implies 3^\nu-1\le 10\implies \nu\le 2$ and trying gives $(m,n)=(4,2)$.