In $x^2(y-2)-x(y-4)=2^{z+1}-2$ for a given $y$, how many solutions $x,z$ are possible?

Question

In $$x^2(y-2)-x(y-4)=2^{z+1}-2$$for a given $y$, how many solutions $x,z$ are possible? I know that there is a finite amount of them, but is there any way to get exactly how many? Also $x,y,z$ are positive integers.

Answer 1

$x^2(y-2)-x(y-4)=2^{z+1}-2 \Rightarrow x^2(y-2)-x(y-4)+(-2^{z+1}+2)=0$

So we have $$x=\frac{(y-4) \pm \sqrt{(y-4)^2+8(y-2)(2^z-1)}}{2(y-2)}$$

For a fixed $y$, you can get two $x$ for each $z$ that you get.