Substitute $2x+a$ in the second equation. After some simplification, you get:
Now write $x=\dfrac pq$ in irreducible form, $q>0$. The equation is equivalent to:
Since $\gcd(p,q)=1$, hence $\gcd(q,p+aq)=1$, $q$ divides $3$.
But it is impossible that $q=3$ since this would imply $ p(p+aq)=3(b-a^2)$, and $3$ cannot divide $p$ nor $p+3a$.
This proves $x$, hence $y$, is an integer.
Alternative proof (by courtesy of PM 2Ring):
$q^2$ is coprime with $p$ and $p+aq$, hence by Gauß's lemma, it divides $3$. However the only square that divides $3$ is $1$, so that $x$ is an integer, whence $y$ is too.