II. (16 points) Given real numbers $x, y$ satisfy
$$
2^{x}+3^{y}=4^{x}+9^{y} \text {. }
$$

Try to find the range of values for $U=8^{x}+27^{y}$.
Let $a=2^{x}, b=3^{y}$.
Then the given condition becomes
$$
\begin{array}{l}
a+b=a^{2}+b^{2} \quad (a, b > 0) \\
\Rightarrow\left(a-\frac{1}{2}\right)^{2}+\left(b-\frac{1}{2}\right)^{2}=\frac{1}{2} .
\end{array}
$$

From the image in the $a O b$ plane, we know
$$
t=a+b \in(1,2] \text {. }
$$
$$
\begin{array}{l}
\text { Also, } a b=\frac{(a+b)^{2}-\left(a^{2}+b^{2}\right)}{2}=\frac{t^{2}-t}{2}, \text { so } \\
U=8^{x}+27^{x}=a^{3}+b^{3} \\
=(a+b)^{3}-3 a b(a+b) \\
=t^{3}-3 \cdot \frac{t^{2}-t}{2} \cdot t=-\frac{1}{2} t^{3}+\frac{3}{2} t^{2} .
\end{array}
$$

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

Thus, $f(t)$ is monotonically increasing on the interval $(1,2]$. It is easy to see that when $t \in(1,2]$, $f(t) \in(1,2]$.
In conclusion, the range of $U=8^{x}+27^{y}$ is $(1,2]$.