If $r = 2$ we get $p^4 + 2p + q^4 + q^2 = 4q^3 + 5$. Then $q = 2, 3$ otherwise $q \ge 5$ and $q^4 + q^2 > 4q^3 + 5$. If $q = 2$ then $p^4 + 2p = 17$ gives no solution. If $q = 3$ we get $p^4 + 2p = 23$ gives no solution. Therefore, $r$ is odd and the right hand side of the main equation is even. Then since $2p + q^4 + q^2 = 2p + q^2(q^2 + 1)$ is even we get that $p^4$, equivalently $p$ is even: $p = 2$. Thus, the main equation becomes $r^2 = q^4 + q^2 - 4q^3 + 19$.

Now we show that there is no solution if $q \ne 2, 3, 5$ since $r^2$ lies between two consecutive perfect squares:
$$(q^2 - 2q - 2)^2 < r^2 < (q^2 - 2q - 1)^2$$
Indeed, $(q^2 - 2q - 2)^2 < r^2 = q^4 + q^2 - 4q^3 + 19$ is equivalent to $(q - 5)(q - 3) > 0$ and $r^2 = q^4 + q^2 - 4q^3 + 19 < (q^2 - 2q - 1)^2$ is equivalent to $q^2 + 4q > 18$. Finally, $q = 2$ gives $r^2 = 7$, $q = 3$ gives $r^2 = 1$, $q = 5$ gives $r = 13$. The only solution is $(2, 5, 13)$.