The equality can be written $p+q = 2(p-q)^2$, which shows that $p$ is odd.
If $p \ge 5$, then $p$ and $q$ leave remainder 1 or 2 when divided by 3.
We will show that in this case the equality is impossible. Indeed, if $p$ and $q$ leave the same remainder mod 3, then $3 \mid 2(p-q)^2$ and $3 \nmid p+q$; if $p$ and $q$ leave different remainders, then $3 \nmid 2(p-q)^2$ and $3 \mid p+q$.
Finally, if $p=3$, then $q=5$.