Junior Macedonian Mathematical Olympiad

It cannot be that $q-p=1$, since $(q+p)^p > 1$, for all prime numbers $p$ and $q$.
Let $r$ be a prime divisor of $q-p$. Then $r$ is also a divisor of $q+p$, so it is a divisor of $2q = (q+p)+(q-p)$ and of $2p = (q+p)-(q-p)$. It follows that $p=q=r$ or $r=2$. The case $p=q$ is impossible, since the right-hand side will be zero, but not the left-hand side. According to this it has to be that $q-p=2^k$ and $q+p=2^l$ for some positive integers $k$ and $l$. From $q = \frac{(q+p)+(q-p)}{2} = 2^{l-1} + 2^{k-1}$, it follows that for $k>1$, $2|q$, which is possible only for $q=2$, but then $(q-p)^{2q-1} < 0 < (p+q)^p$, so it has to be that $k=1$. This means that $q-p=2$. On the other hand we get $lp = 2q-1 = 2p+3$, hence $3|p$, i.e. $p=3$ and $q=5$. By checking we obtain that these numbers satisfy the equation.