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
$$\begin{aligned} p^4 + 4 \cdot 2^{4k} &= p^4 + 4 \cdot 2^{4k} + 4 \cdot 2^{2k} p^2 - 4 \cdot 2^{2k} p^2 = (p^2 + 2 \cdot 2^{2k})^2 - 4 \cdot 2^{2k} p^2 = \\ &= (p^2 + 2 \cdot 2^{2k} + 2 \cdot 2^k p)(p^2 + 2 \cdot 2^{2k} - 2 \cdot 2^k p) = \\ &= (p^2 + 2 \cdot 2^{2k} + 2 \cdot 2^k p)((p - 2^k)^2 + 2^{2k}) \end{aligned}$$
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.