Example 5 There are $n$ people, and it is known that any 2 of them make at most one phone call. The total number of phone calls made among any $n-2$ of them is equal, and it is $3^{k}$ ($k$ is a positive integer). Find all possible values of $n$.

 untranslated text remains the same in terms of line breaks and formatting.
Let the total number of calls among $n$ people be $m$. Since $n$ people can form $C_{n}^{n-2}$ groups of $n-2$ people, and the total number of calls among each $n-2$ people is $3^{k}$, the total number of calls in all $n-2$ groups is $C_{n}^{n-2} \cdot 3^{k}$.

On the other hand, in the above counting, each pair of people who call each other belongs to $C_{n-2}^{n-4}$ groups of $n-2$ people, so each call between 2 people is counted $C_{n-2}^{n-4}$ times. Therefore,
$$
m=\frac{C_{n}^{n-2} 3^{k}}{C_{n-2}^{n-4}}=\frac{n(n-1) \cdot 3^{k}}{(n-2)(n-3)} .
$$
(1) If 3 does not divide $n$, i.e., $(3, n)=1$, then $(n-3, n)=1,\left(n-3,3^{k}\right)=1$.

Also, $(n-2, n-1)=1$, so $n-3 \mid n-1$, i.e.,
$\frac{n-1}{n-3}=1+\frac{2}{n-3}$ is a positive integer, so $n-3 \mid 2, n-3 \leqslant 2, n \leqslant 5$.
Also, $C_{n-2}^{2} \geqslant 3^{k} \geqslant 3, \therefore n \geqslant 5$, hence $n=5$.
(2) If 3 divides $n$, then $3 \mid n-3,3 \times n-2$, i.e., $(3, n-2)=1$. Also, $(n-2, n-1)=1$,

so $n-2 \mid n$, i.e., $\frac{n}{n-2}=1+\frac{2}{n-2}$ is a positive integer,
so $n-2 \leqslant 2$, hence $n \leqslant 4$, which contradicts $n \geqslant 5$.
From (1) and (2), we know that $n$ can only be 5. On the other hand, if there are $n=5$ people, and each 2 people call each other once, then the number of calls among any $n-2=3$ people is $C_{3}^{2}=3^{1}$ (here $k=1$ is a positive integer), which satisfies the problem's requirements. Therefore, the only positive integer is: $n=5$.