VII OBM

If $a^2-4b=c^2-4d$, then $a^2\equiv c^2(\mathrm{mod}4)$, so $a$ and $c$ have the same parity. So if we take any integer $x$ and then $y$ to be the integer $x-\frac{a-c}{2}$ we have $x-\frac{a}{2}=y-\frac{c}{2}$ and hence $(x-\frac{a}{2}{)}^2-\frac{a^2-4b}{4}=(y-\frac{c}{2}{)}^2-\frac{c^2-4d}{4}⟺x^2+ax+b=y^2+cy+d$. Thus the equation has infinitely many integer solutions.

Conversely, suppose $k=(a^2-4b)-(c^2-4d)\ne 0$, then we have $(2x-a{)}^2-(2y-c{)}^2=k$, so $(2x+2y-a-c)(2x-2y-a+c)=k$. But $k$ has only finitely many factorizations, so there are only finitely many possible values for the pair $(2x+2y-a-c,2x-2y-a+c)$ and hence for the pair $(x,y)$.