This is equivalent to $$p+q=2(q-p)^2.$$Note that $$p+q>0\implies q-p\neq 0\implies q>p\ge 2.$$Now, $$2(q-p) ^2\equiv 0 \pmod 2\implies p+q\equiv 0\pmod 2\implies p, q\;\text{are of same parity.}$$Therefore, let $q=p+2k,$ then $$2p+2k=8k^2\implies p=4k^2-k=k(4k-1).$$But since $p$ is a prime, we must have $$k=1.$$Thus, $$q=4k^2+k=5.$$Hence, $(p, q) =(3, 5)$ is the only solution.