$$p^2+5pq+4q^2=k^2 \Leftrightarrow (k-p-2q)(k+p+2q)=pq$$Since $k+p+2q>p,q$ we have $k-p-2q=1$ and $k+p+2q=pq$. Plugging $k=1+p+2q$ into the second equation yields $1+2p+4q=pq \Leftrightarrow (p-4)(q-2)=9$, so $(p,q)=(13,3),(7,5),(5,11)$.