Example 4 Let $m$ and $n$ be positive integers, and the product of all positive divisors of $m$ equals the product of all positive divisors of $n$. Ask: Must $m$ and $n$ be equal?
Solve $m$ and $n$ must be equal.
In fact, by pairing the positive divisor $d$ of $m$ with $\frac{m}{d}$, we know that the product of all positive divisors of $m$ is $m^{\frac{d(m)}{2}}$, thus, the condition is equivalent to
$$m^{d(m)}=n^{d(n)},$$

This equation indicates that $m$ and $n$ have the same prime factors. We can assume
$$m=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}, n=p_{1}^{\beta_{1}} p_{2}^{\beta_{2}} \cdots p_{k}^{\beta_{k}},$$

where $p_{1}d(n)$, then for $1 \leqslant i \leqslant k$, we have $\alpha_{i}<\beta_{i}$, thus, $\alpha_{i}+1<\beta_{i}+1$, hence $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(\alpha_{k}+1\right)<\left(\beta_{1}+1\right)\left(\beta_{2}+1\right) \cdots\left(\beta_{k}+1\right)$, which leads to $d(m)<d(n)$, a contradiction. Similarly, from $d(m)<d(n)$, using (2) can also lead to a contradiction. Therefore, $d(m)=d(n)$, and by (1) we get $m=n$.