jmc

If we let $y=z^2$, then our equation becomes a simple quadratic equation: $$y^2-4y+3=0.$$Indeed, this equation factors easily as $(y-3)(y-1)=0$, so either $y-3=0$ or $y-1=0$.

We now explore both possibilities.

If $y-3=0$, then $y=3$, so $z^2=3$, so $z=\pm \sqrt{3}$.

If $y-1=0$, then $y=1$, so $z^2=1$, so $z=\pm 1$.

Thus we have four solutions to the original equation: $z=\boxed{-\sqrt{3},-1,1,\sqrt{3}}$.