Let $p^2 + 7^n = m^2$. Then $7^n = m^2 - p^2 = (m - p)(m + p)$. We consider two possible cases.

If $m - p = 1$ and $m + p = 7^n$, then $2p = 7^n - 1$. Assume that $n \ge 2$. Then $2p = 7^n - 1 = (7 - 1)(7^{n-1} + 7^{n-2} + \dots + 7 + 1)$, so $p = 3(7^{n-1} + 7^{n-2} + \dots + 7 + 1)$, but this is not possible since the expression in brackets is greater than 1. If $n = 1$, we get $p = 3$.

In the second case we have $m - p = 7^k$ and $m + p = 7^{n-k}$, where $k$ is a positive integer. Obviously $n - k > k$. Hence $2p = 7^{n-k} - 7^k = 7^k(7^{n-2k} - 1)$.

The expression in brackets is even and greater than 2, so $2p$ is the product of at least three primes. This is not possible.

We conclude that $n = 1$ is the only positive integer for which there exists a prime $p$ such that $p^2 + 7^n$ is a perfect square. This prime is $p = 3$ and the perfect square is $3^2 + 7 = 16 = 4^2$.