$x=a^4-6a^2b^2+b^4, \ y=4ab(a^2-b^2), \ z=a^2+b^2.$ you don't need to worry about positivity of $x,y,z$ because if $(x,y,z)$ is a solution, then $(\pm x, \pm y, \pm z)$ will be a solution too.
in order to make sure that $\gcd(x,y,z)=1,$ you should choose $a,b$ so that $\gcd(a,b)=1$ and not both $a,b$ are odd.