Three. (25 points) Let $n$ be a positive integer, and $1 \times 2 \times \cdots \times n$ be denoted as $n!$. Find all positive integer solutions $(x, y)$ to the equation
$$
20(x!)+2(y!)=(2 x+y)!
$$
Three, (1) If $x=y$, then
$$
\begin{array}{l}
22(x!)=(3 x)! \Rightarrow 11 \mid (3 x)! \\
\Rightarrow 3 x \geqslant 11 \Rightarrow x \geqslant \frac{11}{3} \Rightarrow x \geqslant 4 . \\
\text { By } 22(x!)=(3 x)!\geqslant 3 x(3 x-1)(x!) \\
\Rightarrow 9 x^{2}-3 x-22y$, then
$$
\begin{array}{l}
22(x!)>20(x!)+2(y!)=(2 x+y)! \\
\geqslant(2 x+1)!\geqslant(2 x+1) 2 x(x!) \\
\Rightarrow 2 x^{2}+x-11y \geqslant 1$, so, $x=2$.
By $x>y \geqslant 1$, we get $y=1$. After verification, it does not meet the requirements.
(3) If $x20(x!)+2(y!)=(2 x+y)! \\
\geqslant(y+2)! \\
\Rightarrow y^{2}+3 y-20x \geqslant 1$, so, $y=2,3$.
By $y>x \geqslant 1$, when $y=2$, $x=1$ meets the requirements, $(x, y)=(1,2)$;
When $y=3$, $x=1$ or 2, neither meets the requirements.
In summary, all positive integer solutions $(x, y)=(1,2)$.