One, (20 points) Find all integer pairs $(x, y) (x > y > 2012)$ that satisfy
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=\frac{1}{2012},
$$

and $x-y$ is maximized.
$$
\begin{array}{l}
x y-2012 x-2012 y=2012 \\
\Rightarrow(x-2012)(y-2012)=2012 \times 2013 .
\end{array}
$$

From $x-y$ being maximum, we know
$$
(x-2012)-(y-2012)
$$

is maximum. Therefore,
$$
\begin{array}{l}
\left\{\begin{array}{l}
x-2012=2012 \times 2013 \\
y-2012=1
\end{array}\right. \\
\Rightarrow(x, y)=(4052168,2013) .
\end{array}
$$