Assume that $ p^{q-1} -q^{p-1} =4n^{3}$ for distinct odd primes $ p$, $ q$. Note that $ \left( p^{( q-1) /2} +q^{( p-1) /2}\right)\left( p^{( q-1) /2} -q^{( p-1) /2}\right) =4n^{3}$. Since $ \gcd\left( p^{( q-1) /2} +q^{( p-1) /2} ,p^{( q-1) /2} -q^{( p-1) /2}\right) =2$, we can take $ ( A,B) \in \mathbb{N}^{2}$ with $ \gcd( A,B) =1$ such that $ p^{( q-1) /2} +q^{( p-1) /2} =2A^{3}$ and $ p^{( q-1) /2} -q^{( p-1) /2} =2B^{3} $. So we have $ p^{( q-1) /2} =( A+B)\left( A^{2} -AB+B^{2}\right) $. Let $ d:=\gcd\left( A+B,A^{2} -AB+B^{2}\right) =\gcd\left( A+B,3A^{2}\right) =\gcd( A+B,3)$. We must have $ d=3$ otherwise $ ( A+B)\left( A^{2} -AB+B^{2}\right)$ has at least two distinct prime divisors. Note that $ A+B\leq A^{2} -AB+B^{2}$ because $ A\geq 2$ and $ B\geq 1$. So we must have $ A+B=3$, i.e. $ A=2$ and $ B=1$. So we have $ p^{( q-1) /2} +q^{( p-1) /2} =16$ and $ p^{( q-1) /2} -q^{( p-1) /2} =2$. It follows that $ p^{( q-1) /2} =9$ and $ q^{( p-1) /2} =7$, impossible. $ \blacksquare $