Assume that $q_{1}^{4} + q_{2}^{4} + q_{3}^{4} + q_{4}^{4} + q_{5}^{4} = 2k(2k+2)$ for some positive integer $k$. That is
$$q_{1}^{4} + q_{2}^{4} + q_{3}^{4} + q_{4}^{4} + q_{5}^{4} = 4k(k+1)$$
If $p$ is a prime, $p \neq 2$, then $p^{4} \equiv 1 \pmod{8}$.
Assume that $q_{1} \leq q_{2} \leq q_{3} \leq q_{4} \leq q_{5}$. A simple parity argument shows that $q_{1} = 2$. If $q_{2} \neq 2$, then we have
$$q_{1}^{4} + q_{2}^{4} + q_{3}^{4} + q_{4}^{4} + q_{5}^{4} \equiv 4 \pmod{8}$$
Since $8 \mid 4k(k+1) - 16$, we get a contradiction. Therefore $q_{2} = 2$. In similar way it follows $q_{3} = q_{4} = q_{5} = 2$.
The only solution is $q_{1} = q_{2} = q_{3} = q_{4} = q_{5} = 2$, and $k = 4$.