Consider the quadratic $k^2+ak+b$ with real roots $x,y$. Then, by Vieta's, we have $a=-2$ and $b=z^2+1$. So, the roots are $k=\frac{2\pm \sqrt{4-4z^2-4}}{2}=\frac{2\pm \sqrt{-4z^2}}{2}$. In order for $x,y$ to be real, we need $-4z^2\ge 0\iff z^2\leq 0\implies z=0$. Then, plugging into the Quadratic Formula, we get $x,y=1$. So, $x^2+y^2+z^2=\boxed{2}$.