10. (20 points) Given real numbers $x, y$ satisfy
$$
2^{x}+2^{y}=4^{x}+4^{y} \text {. }
$$

Try to find the range of values for $U=8^{x}+8^{y}$.
10. Let $a=2^{x}, b=2^{y}$. Then the given condition can be rewritten as
$$
a+b=a^{2}+b^{2} \quad (a, b>0).
$$

Completing the square, we get
$$
\left(a-\frac{1}{2}\right)^{2}+\left(b-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}.
$$

Observing the image of $(a, b)$ satisfying equation (1) in the rectangular coordinate system ${ }_{a} O b$, it is easy to see that
$$
t=a+b \in(1,2].
$$

Also, note that
$$
a b=\frac{(a+b)^{2}-\left(a^{2}+b^{2}\right)}{2}=\frac{t^{2}-t}{2}.
$$

Therefore, $U=8^{x}+8^{y}=a^{3}+b^{3}$
$$
\begin{array}{l}
=(a+b)^{3}-3 a b(a+b) \\
=t^{3}-3 \times \frac{t^{2}-t}{2} \cdot t=-\frac{1}{2} t^{3}+\frac{3}{2} t^{2}.
\end{array}
$$

Let $f(t)=-\frac{1}{2} t^{3}+\frac{3}{2} t^{2}$. Then, when $t \in(1,2]$,
$$
f^{\prime}(t)=-\frac{3}{2} t^{2}+3 t=-\frac{3}{2} t(t-2)>0.
$$

Thus, $f(t)$ is monotonically increasing on $t \in(1,2]$. It is easy to see that when $t \in(1,2]$, $f(t) \in(1,2]$.
In conclusion, the range of $U=8^{x}+8^{y}$ is $(1,2]$.