JBMO Shortlist

Solution:
By Fermat's theorem $n^{2p} \equiv n^2 \pmod{p}$, therefore $m^2 + n^2 + p + 1 \equiv n^2 \pmod{p} \Rightarrow m^2 \equiv -1 \pmod{p}$.

Case 1: $p = 4k + 3$. We have $(m^2)^{2k+1} \equiv (-1)^{2k+1} \pmod{p}$. Therefore,
$$m^{p-1} \equiv -1 \pmod{p}$$
and $p$ does not divide $m$. On the other hand, by Fermat's theorem
$$m^{p-1} \equiv 1 \pmod{p}$$
(28) and (29) yield $p = 2$. Thus, $p \neq 4k + 3$.

Case 2: $p = 4k + 1$. Let us consider (27) in mod 4. $n^2 = 0$ or $1$ in mod 4. In both cases $n^{2p} = n^2 \pmod{4}$. From (27) we get $n^2 \equiv m^2 + n^2 + 1 + 1 \pmod{4}$. Therefore, $m^2 \equiv -2 \pmod{4}$, and again there is no solution.

Case 3: $p = 2$. The given equation is written as
$$n^4 - n^2 - 3 = m^2$$
Let $l = n^2$. Readily, we do not get any solution for $l = 0$, $1$. If $l = 4$, then there are four solutions: $(3,2,2), (-3,2,2), (3,-2,2), (-3,-2,2)$. There is no solution for $l > 4$, since in this case
$$(l-1)^2 = l^2 - 2l + 1 < m^2 = l^2 - l - 3 < l^2$$
Thus, (27) has four solutions: $(3,2,2), (-3,2,2), (3,-2,2), (-3,-2,2)$ and we are done.