Example 6 Let $k$ be a given positive integer. Try to find all positive integers $a$, such that there exists a positive integer $n$, satisfying $n^{2} \mid \left(a^{n}-1\right)$ and $n$ has exactly $k$ distinct prime factors.
When $a=1$, let
$n=p_{1} p_{2} \cdots p_{k}\left(p_{1}<p_{2}<\cdots<p_{k}\right)$, then $n \nmid\left(2^{n}-1\right)$.
Proof by contradiction.
If $n \mid\left(2^{n}-1\right)$, then $n$ must be odd.
Let $q$ be the smallest prime factor of $n$. Then $q \mid\left(2^{n}-1\right)$ (since $q$ is odd).
By Fermat's Little Theorem, $q \mid\left(2^{q-1}-1\right)$. Therefore, $q \mid\left(2^{(n, q-1)}-1\right) \Rightarrow q \mid\left(2^{1}-1\right) \Rightarrow q \mid 1$. Noting that, by the minimality of $q$ and $q \geqslant 3$, a contradiction arises. Next, we prove: when $a \geqslant 3$, the requirement is always satisfied.
Fix $a$, and prove the strengthened proposition by induction on $k$:
There exists a positive integer $n$, $n$ has exactly $k$ distinct prime factors, each with a power of 1 in $n$.

When $k=1$, take $n$ as a prime factor $p$ of $a-1$, let $a-1=k p\left(k \in \mathbf{Z}_{+}\right)$. Then
$$
\begin{array}{l}
a^{p}-1=(k p+1)^{p}-1 \\
=\sum_{i=2}^{p} \mathrm{C}_{p}^{i}(k p)^{i}+k p^{2} \equiv 0\left(\bmod p^{2}\right),
\end{array}
$$

satisfying the requirement.
Assume that when $k=s$, there exists
$n=p_{1} p_{2} \cdots p_{s}\left(p_{1}<p_{2}<\cdots<p_{s}\right.$ are primes $)$, and $n^{2} \mid\left(a^{n}-1\right)$.

Then by Zsigmondy's Theorem, $a^{n}-1$ has a prime factor $r$, such that for any $1 \leqslant k<n\left(k \in \mathbf{Z}_{+}\right)$, $r \nmid\left(a^{k}-1\right)$.

Thus, for any $i(i=1,2, \cdots, s)$, $r \neq p_{i}$; otherwise, $r=p_{i}, p_{i} \mid\left(a^{p_{i}-1}-1\right)$, and $p_{i}-1<p_{i} \leqslant n$, a contradiction.
Let $n^{\prime}=n r$, then
$a^{n^{\prime}}-1=a^{n r}-1=\left(a^{n}-1\right) \sum_{k=0}^{r-1}\left(a^{n}\right)^{k}$.
By $a^{n} \equiv 1(\bmod r)$, we have
$\sum_{k=0}^{r-1}\left(a^{n}\right)^{k} \equiv 0(\bmod r)$.
Thus, $r^{2} \mid\left(a^{n^{\prime}}-1\right)$.
Since $n^{2} \mid\left(a^{n}-1\right)$ and $\left(a^{n}-1\right) \mid\left(a^{n^{2}}-1\right)$, and $(r, n)=1$, it follows that $\left(n^{\prime}\right)^{2} \mid\left(a^{n^{\prime}}-1\right)$.
This shows that the strengthened proposition holds for $s+1$.
Therefore, for any positive integer $k$, such an $n$ exists.
In summary, the required $a$ is all positive integers not equal to 2.