Parametric characterization for $x^2 + y^2 = 2z^2$

Existence of such integers $a,b$ is equivalent to saying that $x,y$ have the same parity (do you see why?). If $x,y$ had opposite parity, then $x^2+y^2$ would be odd, while $2z^2$ is even.

Now if we have such $a,b$ then we can substitute to get $(a+b)^2+(a-b)^2=2z^2$. Simple algebra gives us that this is equivalent to $a^2+b^2=z^2$. I suspect that now you know the parametrization of solutions to this equation (you also have to show that $a,b,z$ are relatively prime; I leave it to you).

As you have asked for this, here is how to continue: because $(a,b,z)$ is a primitive Pythagorean triple, we have, for some integers $m,n$, that $a=m^2-n^2,b=2mn,z=m^2+n^2$. Now we have $x=a+b=m^2+2mn-n^2,y=a-b=m^2-2mn-n^2$. So the complete parametrization is:

$$(m^2+2mn-n^2,m^2-2mn-n^2,m^2+n^2)$$