@MaryStar dividing by $z^2$, this is equivalent to finding rational solutions to $a^2+3b^2=1$, consider the extension $\Bbb{Q}(\sqrt{-3})/\Bbb{Q}$, note that the norm of $a+b\sqrt{-3}$ is $a^2+3b^2=1$, so we're actually solving the norm equation $N_{\Bbb{Q}(\sqrt{-3})/\Bbb{Q}}(a+b\sqrt{-3})=1$. By Hilbert 90, any solution may be written as $a+b\sqrt{-3}=\frac{c+\sqrt{-3}d}{c-\sqrt{-3}d}$ where we can assume that $c,d \in \Bbb Z$ and $\gcd{c,d}=1$.
One gets $\frac{c+\sqrt{-3}d}{c-\sqrt{-3}d}=\frac{c^2-3d^2+2cd\sqrt{-3}}{c^2+3d^2}$