A solution can be seen here: https://artofproblemsolving.com/community/c4h1785305

Answer: $(p,q) = (1,1); (2,2)$ Solution: Case 1: $q \leq p-1$ $(p+1)^{p-1}+(p-1)^{p+1}=q^q \leq (p-1)^{p-1}$ Clearly there is no solution, since $(p+1)^{p-1}>(p-1)^{p-1}$. Case 2: $q \geq p+2$ $(p+1)^{p-1}+(p-1)^{p+1}=q^q \geq (p+2)^{p+2} > 2(p+2)^{p+1}$ There is no solution for this case, too, because $(p+2)^{p+1}>(p+1)^{p-1}$ and $(p+2)^{p+1}>(p-1)^{p+1}$ $ \implies $ $2(p+2)^{p+1}>(p+1)^{p-1}+(p-1)^{p+1}$, contradiction. Case 3: $q=p+1$ $(p+1)^{p-1}+(p-1)^{p+1}=(p+1)^{p+1}$ $(p-1)^{p+1}=(p+1)^{p-1}(p^2+2p)$, we know that $(p+1)^{p-1}>(p-1)^{p-1}$ and $p^2+2p>(p-1)^2$, multiplying these inequalities we get that $(p+1)^{p-1}(p^2+2p)>(p-1)^{p+1}$, hence no solution either for ths case. Case 4: $q=p$ $(p+1)^{p-1}+(p-1)^{p+1}=p^p$, With a simple induction we can show that $(p-1)^{p+1}>p^p$ for all $p\geq 6$. Base case is true. We must prove that if $(p-1)^{p+1}>p^p$, then $p^{p+2}>(p+1)^{p+1}$. It suffices to show that $p^{p+2}*p^p>(p+1)^{p+1}*(p-1)^{p+1}$ $ \implies $ $p^2>(p-1)(p+1)$, which is clearly true. $\blacksquare$ By checking the values $p \leq 5$, we conclude that $p=1$ and $p=2$ are the only solutions.