II. (40 points) Find all positive integer solutions $(x, y)$ of the equation $x! + y! = x^{y}$.

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II. (40 points) Find all positive integer solutions $(x, y)$ of the equation $x! + y! = x^{y}$.
II. It is known that $x \geqslant 2$.
(1) If $x=y$, then $2 \cdot x!=x^{x}$. Clearly, $(2,2)$ is a solution to the equation.
By mathematical induction, it is easy to prove: when $x \geqslant 3$, $2 \cdot x! < x^x$.
(2) If $x>y$, then $y!\left(1+\frac{x!}{y!}\right)=x^{y}$.
Thus, $\left.\left(1+\frac{x!}{y!}\right) \right\rvert\, x^{y}$.
By $\left(1+\frac{x!}{y!}, x\right)=1$, we get $\left(1+\frac{x!}{y!}, x^{y}\right)=1$.
Thus, $1+\frac{x!}{y!}=1$, which is a contradiction.
(3) If $x<y$, then
$$
\begin{array}{l}
x!\left(1+\frac{y!}{x!}\right)=x^{y} \\
\Rightarrow(x-1)!\left(1+\frac{y!}{x!}\right)=x^{y-1} \\
\Rightarrow(x-1) \mid x^{y-1} . \\
\text { By }(x, x-1)=1 \Rightarrow\left(x-1, x^{y-1}\right)=1 \\
\Rightarrow x-1=1 \Rightarrow x=2 \Rightarrow 2+y!=2^{y} .
\end{array}
$$

If $y \geqslant 4$, then
$$
2^{y} \equiv 2+y!\equiv 2(\bmod 4),
$$

which is a contradiction.
Therefore, $y=3$.
Upon verification, $(2,3)$ is a solution to the original equation.
In summary, $(2,2)$ and $(2,3)$ are all the positive integer solutions to the original equation.