Proof
First, we prove that there exists a pair $(p, q)$ that satisfies the condition.
Consider the following number table:
$$\begin{array}{l} x_{0,1}, x_{0,2}, \cdots, x_{0, m} \\ x_{1,1}, x_{1,2}, \cdots, x_{1, m} \\ \cdots \\ x_{m, 1}, x_{m, 2}, \cdots, x_{m, m} \end{array}$$

where $x_{0,1}=a_{1}, x_{0, j} \rightleftharpoons x_{0, j-1}+1, j=2,3, \cdots, m$. And
$$x_{i, j}=\left(\prod_{k=1}^{m} x_{i-1, k}\right)+x_{i-1, j}, 1 \leqslant i, j \leqslant m$$

The numbers in the above table have the following properties: each row from left to right is exactly a sequence of $m$ consecutive positive integers; in any two numbers $a, b$ in each column, the smaller number (let it be $a$) is a divisor of the larger number, i.e., $a \mid b$.

Since $0 < a_{n+1} - a_{n} \leqslant m$, each row of the table contains at least two terms from $\{a_{n}\}$, so the table contains at least $2(m+1)$ numbers $\in \{a_{n} \mid n=1,2, \cdots\}$. Therefore, there must be a column in the table that contains two numbers simultaneously belonging to $\{a_{n} \mid n=1,2, \cdots\}$, denoted as $a_{p}, a_{q} (p<q)$, then $a_{p} \mid a_{q}$. Thus, we have found a pair $(p, q)$ that satisfies the condition.

Now, set $x_{0,1}$ to $a_{q}+1$, and construct a number table with the same properties. We can find another pair of numbers $\left(p^{\prime}, q^{\prime}\right)$ that satisfy the condition. By continuing this process, we can find infinitely many pairs $(p, q)$ such that $p<q$ and $a_{p} \mid a_{q}$.

The proposition is proved.