, 2016

There is no such triple.
First note that
$$2^m p^2 = q^7 - 1 = (q-1)(q^6 + q^5 + q^4 + q^3 + q^2 + q + 1).$$
Since $q^6 + q^5 + q^4 + q^3 + q^2 + q + 1$ is odd and is larger than $1$, we have $q-1 = 2^m$ or $q-1 = 2^m p$. The latter is impossible by Zsigmondy's theorem, since $q^7 - 1$ has a prime divisor not dividing $q-1$ (alternatively, one can obtain a contradiction by using some simple bounds). Thus, we have $q = 2^m + 1$.
If $m$ is odd, then $q = 2^m + 1 \equiv 2 + 1 \equiv 0 \pmod 3$. As $q$ is a prime, we must have $q=3$, and so $m=1$. But then $2p^2+1=2187$ has no solution.
If $m$ is even, let $m = 2k$. The equation becomes $q^7 - (2^k p)^2 = 1$. By Catalan's conjecture, there is no solution.