I claim the only answer is $x=1$ for which $z=1-y$ is always a solution. Considering the equation as a quadratic in $z$, namely $$z^2 - (2xy)z + x^2+y^2-1=0,$$ our goal is to find all $x$ for which $(x^2-1)(y^2-1)$ is a perfect square for any $y$. The case $x=1$ is immediate, so suppose $x\ne 1$. Clearly $x\ne 0$, so $|x|\ge 2$, yielding $x^2-1\ge 3$. Thus there is a prime $p$ such that $v_p(x^2-1)$ is odd since $x^2-1$ cannot be a perfect square. If $p=2$ then $(x^2-1)(y^2-1)$ cannot be a square for any odd $y$. If $p>2$ then choose $y\not\equiv \pm 1\pmod{p}$ (such a residue class exists), for which $v_p((x^2-1)(y^2-1))$ is odd.