$x^2+ax+b=y^2$ $4y^2=4x^2+4ax+4b=(2x+a)^2+4b-a^2$ $(2y-2x-a)(2y+2x+a)=4b-a^2$ As $b$ is not square, than $4b-a^2 \neq 0$ Then $4x+2a=d-\frac{4b-a^2}{d}$ where $d$ is divisor of $4b-a^2$ and so $x$ takes only finite values