II. (40 points) Let $m, n$ be positive integers, and $p$ be a prime. Find all triples $(m, n, p)$ that satisfy $\mathrm{C}_{m}^{3}-4=p^{n}$.
Notice,
$$
\mathrm{C}_{m}^{3}-4=\frac{1}{6}(m-4)\left(m^{2}+m+6\right)=p^{n} \text {. }
$$

Thus, it must be that $m>4$.
$$
\begin{array}{l}
\text { By }\left(m-4, m^{2}+m+6\right)=(m-4,5 m+6) \\
=(m-4,26),
\end{array}
$$

We consider three cases.
(1) $p \neq 2,13$.
Then $\left(m-4, m^{2}+m+6\right)=1$, so one of them must be a divisor of 6.
Since $m^{2}+m+6>6 \Rightarrow(m-4) 16$.
Testing, we find $(m, n, p)=(7,1,31)$.
(2) $p=2$.
In this case, $(m-4)\left(m^{2}+m+6\right)=3 \times 2^{n+1}$.
(i) If $m-4=3 k$, then
$k=2^{s}, m^{2}+m+6=2^{t}$.
Since the common divisor divides 26, one of them must be 2, i.e., $k=2, m=10$, which is not valid.
(ii) If $m^{2}+m+6=3 k$, then $m-4=2^{s}$.
Similarly, $m-4=2 \Rightarrow m=6$.
Solving, we get $(m, n, p)=(6,4,2)$.
(3) $p=13$.
In this case, $(m-4)\left(m^{2}+m+6\right)=6 \times 13^{n}$.
Considering $m-4$ and $m^{2}+m+6$ equal to $13, 26, 39, 78$, none of them are valid.
In summary, $(m, n, p)=(7,1,31),(6,4,2)$.