jmc

The expression $z^2-3z+1$ appears twice in the equation we're trying to solve. This suggests that we should try the substitution $y=z^2-3z+1$. Applying this to the left side of our original equation, we get $$y^2-3y+1=z,$$which, interestingly, looks just like the substitution we made except that the variables are reversed. Thus we have a symmetric system of equations: $$\begin{array}{rl}y& =z^2-3z+1,\\ y^2-3y+1& =z.\end{array}$$Adding these two equations gives us $$y^2-2y+1=z^2-2z+1,$$which looks promising as each side can be factored as a perfect square: $$(y-1{)}^2=(z-1{)}^2.$$It follows that either $y-1=z-1$ (and so $y=z$), or $y-1=-(z-1)$ (and so $y=2-z$). We consider each of these two cases.

If $y=z$, then we have $z=z^2-3z+1$, and so $0=z^2-4z+1$. Solving this quadratic yields $z=\frac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}$.

If $y=2-z$, then we have $2-z=z^2-3z+1$, and so $2=z^2-2z+1=(z-1{)}^2$. Thus we have $z-1=\pm \sqrt{2}$, and $z=1\pm \sqrt{2}$.

Putting our two cases together, we have four solutions in all: $z=\boxed{1+\sqrt{2},1-\sqrt{2},2+\sqrt{3},2-\sqrt{3}}$.