3. For any positive integer $n$, let
$$
P_{n}=1!\cdot 2!\cdots \cdot n!\text {. }
$$
(1) Find all positive integers $m$ such that $\frac{P_{2020}}{m!}$ is a perfect square;
(2) Prove: there exist at least two positive integers $m$ such that there are infinitely many positive integers $n$ satisfying $\frac{P_{n}}{m!}$ is a perfect square.
3. (1) Notice,
$$
\begin{array}{l}
P_{2010}=\prod_{i=1}^{2020} \prod_{j=1}^{i} j=\prod_{j=1}^{2020} \prod_{i=j}^{2020} j=\prod_{j=1}^{2020} j^{2021-j} \\
=\left(\prod_{k=1}^{1009}((2 k)(2 k+1))^{1010-k}\right)^{2} \times \\
2^{1010} \times 1010!.
\end{array}
$$

This indicates that $m=1010$ is a positive integer that satisfies the condition.
Suppose there is another positive integer $m$ that also satisfies the condition.
Since $\frac{P_{2020}}{1010!}$ is a perfect square, we have
$$
\frac{\frac{P_{2020}}{m!}}{\frac{P_{2020}}{1010!}}=\frac{1010!}{m!}
$$

is the square of a rational number.
If $m < 1010$,
$$

then $\frac{1010!}{m!}$ is not a perfect square, which contradicts the assumption.
If $m > 1010$,
$$

then $\frac{1010!}{m!}$ is not a perfect square, which also contradicts the assumption.
Therefore, $\frac{P_{n}}{(m+1)!}$ is a perfect square.