Sketch of the proof: $p$ or $r$ is even, so $2$. If $p=2$, we get $q^4-4q^3+q^2-19$ is a perfect square and it's a well-known technique to look for an $a$ w.t.p. $a^2<q^4-4q^3+q^2-19<(a+1)^2$, and if $r=2$, $p^4+2p+q^4-4q^3+q^2=5$, which clearly doesn't have solutions for $q \ge 4$ and the other cases are easy.

Indeed to this end, we have that for $q>5$ $(q^2-2q-2)^2<q^4-4q^3+q^2+19<(q^2-2q-1)^2.$ And solutions are $q=5,\ r=13$ if I'm not mistaken.

It can also be done by well known property of prime $\ge 5$ that is $p \equiv 1,-1 \pmod{6}$ $\Box$