Example 2 uses $P^{*}$ to denote the set of all odd prime numbers less than 10000. Let $p$ be a number in $P^{*}$, satisfying: for any subset $S=\left\{p_{1}, p_{2}, \cdots, p_{k}\right\}$ of $P^{*}$ that does not contain $p$ and where $k \geqslant 2$, there exists $q \in \mathbf{P}^{*} \backslash S$ such that
$$(q+1) \mid\left(\left(p_{1}+1\right) \cdot\left(p_{2}+1\right) \cdot \cdots \cdot\left(p_{k}+1\right)\right)$$

Find all possible values of $p$.
A basic idea is to minimize the number of distinct prime factors in the numerator of (1), the simplest case being to make each $p_{i}+1$ a power of 2. Therefore, we examine the Mersenne primes in $P^{*}$.

Let $T=\left\{M_{2}, M_{3}, M_{5}, M_{7}, M_{13}\right\}=\{3,7,31,127,8191\}$ (note that $M_{11}=$ $23 \times 89$ is not a prime), then $T$ is the set of all Mersenne primes in $P^{*}$.

Now we discuss all possible values of $p$.
On one hand, if $p \notin T$, then by taking $S=T$ in the condition, there exists $q \in P^{*} \backslash S$ such that $(q+1) \mid\left(M_{2}+1\right)\left(M_{3}+1\right)\left(M_{5}+1\right)\left(M_{7}+1\right)\left(M_{13}+1\right)$, hence $(q+1) \mid 2^{30}$, which indicates that $q+1$ is a power of 2, thus $q \in T$, a contradiction. Therefore, $p \in T$.

On the other hand, for any $p \in T$, we prove that $p$ meets the requirement.
If $p$ does not meet the requirement, there should exist a set $S=\left\{p_{1}, p_{2}, \cdots, p_{k}\right\} \subseteq P^{*}, k \geqslant 2$, not containing $p$, such that any prime $q$ satisfying (1) belongs to $S$. By this property, since $4 \mid\left(p_{1}+1\right)\left(p_{2}+1\right)$, we have $M_{2} \in S$, then by $8 \mid\left(M_{2}+1\right)\left(p_{2}+1\right)$, we have $M_{3} \in S$, by $2^{5} \mid\left(M_{2}+1\right)\left(M_{3}+1\right)$, we know $M_{5} \in S$, by $2^{7} \mid\left(M_{2}+1\right)\left(M_{3}+1\right)\left(M_{5}+1\right)$, we know $M_{7} \in S$, and by $2^{13} \mid\left(M_{2}+1\right)\left(M_{3}+1\right)\left(M_{5}+1\right)\left(M_{7}+1\right)$, we know $M_{13} \in S$. This leads to $T \subseteq S$, contradicting $p \notin S$. Therefore, when $p \in T$, $p$ meets the requirement.
In summary, the $p$ that satisfies the condition is $p \in\left\{M_{2}, M_{3}, M_{5}, M_{7}, M_{13}\right\}$.