Observe that $a_1=p-3q$ and $a_2 = \textstyle 9\binom{q}{2}-3pq+\binom{p}{2}$. From the equality of polynomials, we get $p-3q = \textstyle 9\binom{q}{2}-3pq+\binom{p}{2} \iff (p-3q)^2= 3(p+q)$. Now, we'll do something more: we'll find all solutions to this. Clearly $p+q=3k^2$ for some $k$. Inserting, we get $p-3q=3k$ and $p+q=3k^2$. Thus $4q=3k(k-1)$ and $4p=3k(3k+1)$. If $k$ is even, this amounts to having $k=4\ell$ for some $\ell$, from which we get infinitely many $(p,q)$ pairs. If $k$ is odd, then $k\equiv 1\pmod{4}$. Setting $k=4i+1$, we recover the second family.