The first equation can be transformed as follows:
$$\begin{aligned} & 2^{\sqrt[3]{x^2}} \cdot 4^{\sqrt[3]{y^2}} \cdot 16^{\sqrt[3]{z^2}} = 128 \\ \Leftrightarrow & 2^{\sqrt[3]{x^2} + 2 \cdot \sqrt[3]{y^2} + 4 \cdot \sqrt[3]{z^2}} = 2^7 \\ \Leftrightarrow & \sqrt[3]{x^2} + 2\sqrt[3]{y^2} + 4\sqrt[3]{z^2} = 7 \end{aligned}$$

and the second as
$$\begin{align*} (xy^2 + z^4)^2 &= 4 + (xy^2 - z^4)^2 \\ \Leftrightarrow x^2y^4 + 2xy^2z^4 + z^8 &= 4 + (x^2y^4 - 2xy^2z^4 + z^8) \\ \Leftrightarrow 4xy^2z^4 = 4 \\ \Leftrightarrow xy^2z^4 = 1. \end{align*}$$
Because of the second equation, we note that $x$ must be positive, and we therefore have $x = |x|$.
Since the values of two different means $m_0$ and $m_{2/3}$ of the same variables are equal, each variable must be equal to the mean value 1. We therefore have $x = |y| = |z| = 1$, and the set of solutions of the given system of equations is therefore $(1, -1, -1)$, $(1, -1, 1)$, $(1, 1, -1)$, $(1, 1, 1)$.
qed