For $q = 2$ we obtain $p = 5$, which is a prime. For $q \ge 3$, reducing modulo $4$, we get $p^3 \equiv 3 \pmod 4$, which leads to $p \equiv 3 \pmod 4$. The equation reduces to $p^3 + 125 = 34q^2 + 48q + 18$, or, equivalently,
$$(p+5)(p^2-5p+25) = 2[q^2 + (4q+3)^2].$$
As $p^2 - 5p + 25 \equiv 3 \pmod 4$, it follows that $p^2 - 5p + 25$ must have at least one prime factor $d \equiv 3 \pmod 4$ and, because of $d \mid q^2 + (4q+3)^2$, it follows that $d \mid q$ and $d \mid 4q+3$. In conclusion, $d \mid 3$, which means that $d = q = 3$ and $p = 7$, which is a prime. The solutions to the equation are $(p, q) \in \{(5, 2), (7, 3)\}$.