If we write down the given equation in the form $2^m p^2 = n^5 - 1$ and factorise the right-hand side, we get
$$2^m p^2 = (n-1)(n^4 + n^3 + n^2 + n + 1).$$
Factor $n^4 + n^3 + n^2 + n + 1$ is odd, so $n-1$ is divisible by $2^m$.
We immediately see that $p$ is odd.
On the other hand, since $n$ is positive, we clearly have $n^4 + n^3 + n^2 + n + 1 > n - 1$. Hence $p$ cannot divide $n-1$, because otherwise $n-1$ would be at least $2^m p$, and $n^4 + n^3 + n^2 + n + 1$ would be at most $p$, which is less than $2^m p$. Hence, we have
$$2^m + 1 = n, \quad p^2 = n^4 + n^3 + n^2 + n + 1.$$

$$n^4 + n^3 + n^2 + n = p^2 - 1,$$
$$n(n + 1)(n^2 + 1) = (p - 1)(p + 1).$$
By plugging $n = 2^m + 1$ into the last equation we get
$$(2^m + 1)(2^m + 2)(2^{2m} + 2^{m+1} + 2) = (p - 1)(p + 1),$$
which leads us to
$$4(2^{m-1} + 1)(2^m + 1)(2^{2m-1} + 2^m + 1) = (p - 1)(p + 1).$$
Since $p$ is odd, the right-hand side is the product of two consecutive even numbers, so it is divisible by 8. The left-hand side is not divisible by 8, unless $m = 1$.
It follows that the only solution is $(p, m, n) = (11, 1, 3)$.