16. Given real numbers $x, y$ satisfy $2^{x+1}+2^{y+1}=4^{x}+4^{y}$, find the range of $M=8^{x}+8^{y}$.
Let $u=2^{x}, v=2^{y}$, then $2(u+v)=u^{2}+v^{2} \Rightarrow (u+v)^{2}=u^{2}+v^{2}+2uv$
$$
\Rightarrow 0<2uv=(u+v)^{2}-2(u+v) \leqslant u^{2}+v^{2}=2(u+v) \Rightarrow 2<u+v \leqslant 4 \text{. }
$$

Thus, $u^{3}+v^{3}=(u+v)^{3}-3uv(u+v)=(u+v)^{3}-\frac{3}{2}(u+v)\left[(u+v)^{2}-2(u+v)\right]$
$$
=-\frac{1}{2}(u+v)^{3}+3(u+v)^{2} \xlongequal{t=x+y \in(2,4]}-\frac{1}{2}t^{3}+3t^{2} \text{. }
$$

Let $f(t)=-\frac{1}{2}t^{3}+3t^{2}, f^{\prime}(t)=-\frac{3}{2}t^{2}+6t=\frac{3}{2}t(4-t) \geqslant 0$, thus $f(t)$ is monotonically increasing $\Rightarrow f(t) \in(8,16]$.
Therefore, the range of $M=8^{x}+8^{y}$ is $(8,16]$.