Hint

Answer: $p, q=(3;2), (2;3).$ Case 1: $q=2$. In this case we have $3p-6$ is prime. Since $3|3p-6$, we get $3=3p-6$ and so $p=3$. Case 2: $q>2$. In this case $2|p(q+1)-6=pq+p-6$. Since $pq+p-6$ is prime, we have $2=pq+p-6$ or $8=p(q+1)$. So, $p|8 \Rightarrow p=2 \Rightarrow q=3.$ And we get our answer.

Assume both $p$ and $q$ odd, and note that then $2 \mid pq + q - 6$. This forces $p(q + 1) = 8$ which gives no solutions as $8$ has no odd prime divisors. Thus either $p$ or $q$ is $2$. Say we have $p = 2$, so that we may find, $2q + q - 6 = 3q - 6$ should be prime, which is impossible unless $3q - 6 = 3 \implies q = 3$. Otherwise we have $q = 2$, we may find $2p - 4$ should be prime, which implies $2p - 4 = 2$ or $p = 3$. Then the only solutions are $(2, 3)$ and $(3, 2)$.