The answer is $(m, n, p) = (2, 3, 3)$, $(1, 1, 2)$ or $(2, 2, 5)$.
Let $\frac{5^m + 2^n p}{5^m - 2^n p} = k^2$ for some positive integer $k$. Note that $5^m - 2^n p \mid 5^m + 2^n p$ implies that $5^m - 2^n p \mid 2 \cdot 5^m$. Then as $5^m - 2^n p$ is odd, $5^m - 2^n p \mid 5^m$ and hence $5^m - 2^n p = 5^r$ for some non-negative integer $r$.

Case 1: $r = 0$ i.e. $5^m - 2^n p = 1$.
If $n \ge 3$, then $5^m \equiv 1 \pmod 8$ and hence $m = 2s$ for some positive integer $s$. Then $5^{2s} \equiv 1 \pmod 3$ and we have $2^n p \equiv 0 \pmod 3$. Thus, $p = 3$ and $(5^s - 1)(5^s + 1) = 3 \cdot 2^n$. Observe that $5^s + 1 \equiv 2 \pmod 4$ and has an odd divisor greater than 3 when $s > 1$.

Therefore $s = 1$ and hence $m = 2$, $n = 3$ and $k = 7$.
If $n = 2$, then $8p = (5^m + 2^2p) - (5^m - 2^2p) = k^2 - 1$. Therefore $k = 2l + 1$ for some positive integer $l$ and $2p = l(l+1)$. Then clearly $p = 3$ and hence $5^m = 13$ which yields a contradiction.
If $n = 1$, then $4p = (5^m + 2^1p) - (5^m - 2^1p) = k^2 - 1$. Therefore $k = 2l + 1$ for some positive integer $l$ and $p = l(l+1)$. Then clearly $l = 1$, $p = 2$ and hence $k = 3$, $m = 1$.

Case 2: $r \ge 1$.
Then $5 \mid 2^n p$ and hence $p = 5$. Therefore, $5^{m-1} - 2^n = 5^{r-1}$ implies that $r = 1$ since $m > r$ and $5^{r-1} \mid 2^n$. Thus, we have $5^{m-1} - 2^n = 1$. Clearly $n \ne 1$ and if $n = 2$, then $m = 2$ and $k = 3$.
If $n \ge 3$, then $5^{m-1} \equiv 1 \pmod 8$ and hence $m-1 = 2s$ for some positive integer $s$. Then $(5^s - 1)(5^s + 1) = 2^n$. Observe that $5^s + 1 \equiv 2 \pmod 4$ and has an odd divisor greater than 1 when $s \ge 1$. Therefore $s = 0$ and hence $2^n = 0$ which yields a contradiction.