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{align*} y &= z^2-3z+1, \\ y^2-3y+1 &= z.\end{align*}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} $.