Solve the system of equations in integers:

$$
\left\{\begin{array}{l}
z^{x}=y^{2 x} \\
2^{z}=4^{x} \\
x+y+z=20 .
\end{array}\right.
$$
From the second and third equation we find $z=2 x$ and $x=\frac{20-y}{3}$. Substituting these into the first equation yields $\left(\frac{40-2 y}{3}\right)^{x}=\left(y^{2}\right)^{x}$. As $x \neq 0$ (otherwise we have $0^{0}$ in the first equation which is usually considered undefined) we have $y^{2}= \pm \frac{40-2 y}{3}$ (the ' - ' case occurring only if $x$ is even). The equation $y^{2}=-\frac{40-2 y}{3}$ has no integer solutions; from $y^{2}=\frac{40-2 y}{3}$ we get $y=-4, x=8, z=16$ (the other solution $y=\frac{10}{3}$ is not an integer).

Remark. If we accept the definition $0^{0}=1$, then we get the additional solution $x=0, y=20, z=0$. Defining $0^{0}=0$ gives no additional solution.