Prove that $y^2z^2 - y^2 -z^2$ is not a perfect square for any $y,z \in \mathbf{N}$

Question

So, I was looking at this problem which asks to show that $x^2 + y^2 + z^2 = 2xyz $ where $ x,y,z \in \mathbf{N}$ has no solution. Viewing the equation as quadratic in $x$ and solving for $x$ gives $$ x = yz \pm \sqrt{y^2z^2-y^2-z^2} $$ From which we can conclude $x$ has integral solutions $\iff y^2z^2-y^2-z^2$ is a perfect square. How do I proceed from here to show that $y^2z^2-y^2-z^2$ can never be a perfect square?

Answer 1

Looking at $y^2z^2-y^2-z^2$ modulo $4$, we see that both $y$ and $z$ must be even if that expression is a perfect square. Setting $y=2y_1, z=2z_1$, we get $$ y^2z^2-y^2-z^2=16y_1^2z_1^2-4y_1^2-4z_1^2\\ =4(4y_1^2z_1^2-y_1^2-z_1^2) $$ If this is a perfect square, then $4y_1^2z_1^2-y_1^2-z_1^2$ must also be a perfect square. And again, looking at it modulo $4$, we see that $y_1$ and $z_1$ must both be even.

This continues indefinitely. Which cannot be done with positive integers. So there is no solution (unless $0$ is allowed).