This can be rearranged to $xy(q-p)=pq(x+y)$ which in turn can be rearranged to $((q-p)x-pq)((q-p)y-pq)=p^2q^2$ here we can find solutions of $(x,y)$ as either the first bracket is $1,p,pq,pq^2,p^2,p^2q,p^2q^2$ and same with the second bracket. The cases that $x,y$ are integers are all the pairs that exist satisfying the question.