Remark that $\gcd(5^m + 2^np, 5^m - 2^np) = \gcd(5^m + 2^np, 2 \cdot 5^m) = \gcd(5^m + 2^np, 2^{n+1}p)$, hence their gcd divides $\gcd(2 \cdot 5^m, 2^{n+1}p) = 2,10$ and hence the gcd must be one of $1,2,5,10$. But remark that for $\frac{5^m + 2^np}{5^m-2^np}$ to be an integer we need $5^m-2^np$ to divide the gcd, hence $5^m - 2^np = 1,2,5,10$. It cannot equal one of $2,10$ because that $m,n \ge 1$ but then looking at parity we get a contradiction. If $5^m - 2^np = 1$, then observe that $2^{n+1}p = k^2 - 1 = (k-1)(k+1)$. Remark either $k-1, k+1$ are $2,2^np$ or are $2p, 2^n$ because that from $2 \cdot 5^{m} = k^2 + 1$ we get $k$ is odd. In the first case we get $k=3 \implies p=2, n=1, m=1$ as a solution. In the second case we have either $2^{n-1} - 1 = p$ or $2^{n-1}+1 = p$. When $2^{n-1} - 1 = p$, we have $5^m - 2^n(2^{n-1} - 1) = 1 \implies 5^m - 2^{2n-1} + 2^n = 1$. But then $5^m \equiv 1 \pmod{2^n}$, hence $2^{n-2}|m$. But then $5^m - 2^{2n-1} + 2^n > 1$ for $n \ge 3$, contradiction and hence no solutions in this case. When $2^{n-1}+1 = p$ then $5^m - 2^{2n-1} - 2^n = 1$. By similar argumentation $2^{n-2}|m$, and thus $5^m - 2^{2n-1} - 2^n > 1$ for $n \ge 4$. For $n=3$ we get the solution $n=3, p=3, m=2$. Now suppose $5^m - 2^np = 5$. Clearly then $p=5$, so we get $5^{m-1} - 2^n = 1$. By the Catalan conjecture (or by just using $2^{n-2}|(m-1)$ to eliminate "big" solutions) we get the only solution is $n=2,m=2,p=5$. Hence the only solutions to this are $(n,m,p) = (1,1,2), (3,2,3), (2,2,5)$.

Darn...why such a easy question on tst? especially turkey tst...gcd kills this problem