, 2011

If both $m$ and $n$ are greater than one, then $6^m + 2^n + 2 \equiv_4 0 + 0 + 2 \equiv_4 2$, which is not a perfect square. So at least one of $m$ and $n$ has to be exactly $1$.

* If $m = 1$, then we want $2^n + 8$ to be a square. This has a solution only for $n = 3$. If $n \ge 4$ the expression will be divisible by $8$ and not $16$, so it cannot be a square.

* If $n = 1$, then we want $6^m + 4$ to be a square. Considering this expression mod $7$, we have $(-1)^m + 4$, which will be $3$ or $5$ (mod $7$). However, perfect squares are only $0$, $1$, $2$ or $4$ (mod $7$), so there are no solutions.

So the unique solution to the problem is $(m, n) = (1, 3)$.