jmc

By AM-GM, $$\begin{array}{rl}{x}^4+4y^2+4z^4& =x^4+2y^2+2y^2+4z^4\\ & \ge 4\sqrt[4]{(x^4)(2y^2)(2y^2)(4z^4)}\\ & =8xyz\\ & =16.\end{array}$$Equality occurs when $x^4=2y^2=4z^2.$ Using the condition $xyz=2,$ we can solve to get $x=y=\sqrt{2}$ and $z=1,$ so the minimum value is $\boxed{16}.$