Macedonian Mathematical Olympiad

It is clear that $r>2$, from where $r$ must be an odd prime number. One of the numbers $p$ or $q$ must be $2$, and the other must be an odd prime number. Without loss of generality, let $q=2$ and $p$ be odd. But then the equation is of the form $p^4+2^{2p}=r$, i.e. $p^4+4\cdot 2^{4k}=r$ where $p=2k+1$, $k\in \mathbb{N}$. But then $$\begin{array}{rl}{p}^4+4\cdot 2^{4k}& =p^4+4\cdot 2^{4k}+4\cdot 2^{2k}{p}^2-4\cdot 2^{2k}{p}^2=(p^2+2\cdot 2^{2k}{)}^2-4\cdot 2^{2k}{p}^2=\\ & =(p^2+2\cdot 2^{2k}+2\cdot 2^{k}p)(p^2+2\cdot 2^{2k}-2\cdot 2^{k}p)=\\ & =(p^2+2\cdot 2^{2k}+2\cdot 2^{k}p)((p-2^k{)}^2+2^{2k})\end{array}$$ That means that the number $p^4+2^{2p}$ is never prime, which means that the equation has no solution in the set of prime numbers.