@Kaede That was a very elegant and well thought out proof by contradiction!

A more direct factorisation is $(p^{\frac{q-1}2}+2n)(p^{\frac{q-1}2}-2n)=q^{p-1}$. Clearly $q$ can’t divide both brackets, as then $q \mid p^{\frac{q-1}2}$, a clear contradiction. This now forces $p^{\frac{q-1}2}-2n=1$, which in turn forces $2p^{\frac{q-1}2}-1=q^{p-1}$. But LHS $\equiv -1 \pmod{p}$, whereas RHS $\equiv 1 \pmod{p}$, hence contradiction. Lastly, it was assumed throughout the argument that neither $p$ nor $q$ is even, and that is obvious for parity reasons from the original equation. This problem would be half as difficult if they gave the problem statement as: $p^{q-1}-4n^2=q^{p-1}$.

We first notice that both $p$ and $q$ are odd. Furthermore, $2n, p$ and $q$ are pairwise relatively prime. Then we have $$(2n)^2+(q^{\frac{p-1}{2}})^2=(p^{\frac{q-1}{2}})^2,$$which implies that $(2n, q^{\frac{p-1}{2}}, p^{\frac{q-1}{2}})$ is a primitive Pythagorean triple. This implies that $q^{\frac{p-1}{2}}=a^2-b^2$ and $p^{\frac{q-1}{2}}=a^2+b^2$ for some relatively prime positive integers $a$ and $b$. However, this implies both $a-b$ and $a+b$ are powers of $q$, and if $a-b \neq 1$, then we can easily conclude that $q \mid b$ and $q \mid a$, which is impossible because $a$ and $b$ are relatively prime. Then we must have $a-b=1$ and $a+b=q^{\frac{p-1}{2}}$. Solving the given system and using some algebra, we see that $2p^{\frac{q-1}{2}}=2(a^2+b^2)=q^{p-1}+1$, but the right hand side is not divisible by $p$ by Little Fermat's theorem.

Lemma : $a^2+1$ can't be divisble by $p=4k+3$ $(p^{\frac{q-1}{2}}+q^{\frac{p-1}{2}})(p^{\frac{q-1}{2}}-q^{\frac{p-1}{2}})=4n^2$ we know that $p\ne q$ ,so $(p^{\frac{q-1}{2}}+q^{\frac{p-1}{2}};p^{\frac{q-1}{2}}-q^{\frac{p-1}{2}})=1$, therefore $p^{\frac{q-1}{2}}+q^{\frac{p-1}{2}}=b^2\rightarrow p^{\frac{q-1}{2}}+q^{\frac{p-1}{2}}+1=b^2+1$ if $p,q \ne 2$ ,$p^{\frac{q-1}{2}}+q^{\frac{p-1}{2}}\equiv 2 \pmod 4$ so $b^2+1$ divisble by $p=4k+3$ but it's impossible .