Example 8 Find all positive integers $n$ greater than 1 such that $2^{2}+3^{2}+\cdots+n^{2}$ is a $k$-th power of a prime $p$, where $k$ is a positive integer.
(2004 Polish Mathematical Olympiad)
Analysis We first simplify $2^{2}+3^{2}+\cdots+n^{2}$, and then use the properties of prime numbers.
Solution Since $f(n)=2^{2}+3^{2}+\cdots+n^{2}=\frac{(n-1)\left(2 n^{2}+5 n+6\right)}{6}$,
let $n-1=y$, then $f(n)=\frac{y\left(2 y^{2}+9 y+13\right)}{6}$,
by the problem
$$
6 f(n)=y\left(2 y^{2}+9 y+13\right)=6 p^{k} .
$$
(1) If $p \nmid y$, then $y=1,2,3$ or 6, i.e., $n=2,3,4,7$.
Verification shows that all these satisfy the condition.
(2) If $p \mid y$, since $2 y^{2}+9 y+13>6$, it follows that: $p \mid 2 y^{2}+9 y+13$ which implies $p \mid 13$, so $p=13$, and $13^{2} \mid y$, otherwise since $2 y^{2}+9 y+13>6 \times 13$, it follows that $p^{2} \mid 2 y^{2}+9 y+13$ which implies $p^{2} \mid 13$, a contradiction.
Therefore, $y=13,2 \times 13,3 \times 13,6 \times 13$, and $13^{2} \mid 2 y^{2}+9 y+13$.
However, when $y=13,2 \times 13,3 \times 13,6 \times 13$, calculations show that $9 y+13$ is not divisible by $13^{2}$, i.e., $13^{2} \nmid 2 y^{2}+9 y+13$, a contradiction, so there are no values of $y$ that satisfy the condition in this case.
In summary, $n=2,3,4,7$ are all the values that satisfy the condition.
!Comment The discussion in this problem is based on the relationship between $p$, $y$, and the coefficient 6.