Anti-Pythagorean equation

We have $${1040}^{-2}+{4095}^{-2}={1008}^{-2}$$
How many solutions are there to the "anti-pythagorean" equation $x^{-2}+y^{-2}=z^{-2}$ with $x,y,z\text{ coprime and }x\le y\text{ and }x\le {10}^{12}?$

Answer format: $count,\sum z$

[My timing: 9 sec]