A triple of pythagorean triples with an extra property I'm trying to prove the non-existance of three positive integers $x,y,z$ with $x\geq z$ such that\begin{align}
(x-z)^2+y^2 &\text{ is a perfect square,}\\
x^2+y^2 &\text{ is a perfect square,}\\
(x+z)^2+y^2 &\text{ is a perfect square.}
\end{align}
I failed tying to find such triple numerically. I tried to look at the differences, but that didn't give me much. Also, I tried the general solution of $a^2+b^2=c^2$, that is, $a=2pq$, $b=q^2-p^2$, and $c=q^2+p^2$ for some $p<q$, but this got ugly quite fast. I am not sure whether or not such $x,y,z$ exist. If it exists, one counterexample suffices to disprove my conjecture. Thanks in advance!

It is impossible that non-existance because there are infinitely many counterexamples. In fact, we must have by the Pythagorean triples
$$x-z=t^2-s^2;\space y=2ts\qquad (*)$$ $$x=t_1^2-s_1^2; \space y=2t_1s_1$$  $$x+z=t_2^2-s_2^2;\space \space y=2t_2s_2\qquad (**)$$ so we have from $(*)$ and $(**)$ $$x=\frac{t^2-s^2+t_2^2-s_2^2}{2}=t_1^2-s_1^2\iff t^2+t_2^2-2t_1^2=s^2+s_2^2-2s_1^2$$ On the other hand the identity
$$(m^2-n^2)^2+(2mn)^2=2(m^2+n^2)^2$$ shows infinitely many solutions to the equation $X^2+Y^2=2Z^2$.
Thus we can get an infinite set of possible $t_1,t_2,t_3$
