If $y^2=z^2=c$, and x is any positive integer, then there is a solution.

It's easy to see that $(x, y, z) =(x, x+1, x(x+1)-c)$ is always a solution for the first equation. Plug in the second equation $z=xy-c$. We find that $c=(x-y) ^2/2$. Therefore we must choose $x, y$ to have the same parity and to be distinct. So if $y=x-2$, then $c=x^2-2$ works. Thus, $(x, x-2, 2x-2)$ is a solution for $c=x^2-2$.