Let $x^2=a, y^2=b$. We have \begin{align*}7ab+4a-77b=1260 \\ \implies (a-11)(7b+4)=1216.\end{align*}Because $1216=2^6\cdot 19$, this number only has 14 factors, and we can manually check. We see that the only perfect squares ordered pair of $(a, b)$ is $(49, 4)$, so the integer solutions $(x, y)$ are $\boxed{(7, -2), (-7, 2), (-7, -2), (7, 2)}$.