Example 4 Find all positive integers $m(>1)$, such that the number $m^{3}$ can be expressed as the sum of squares of $m$ consecutive positive integers.
Let $m$ be a positive integer that meets the requirement, then there exists $k \in \mathbf{N}$, such that
$$m^{3}=(k+1)^{2}+(k+2)^{2}+\cdots+(k+m)^{2} .$$
Equation (11) is equivalent to
$$k^{2}+(m+1) k+\frac{1}{6}(m+1)(2 m+1)-m^{2}=0$$

Solving the above quadratic equation in $k$, we get
$$k=-\frac{1}{2}(m+1) \pm \frac{1}{6} t$$

where $t=\sqrt{33 m^{2}+3}$. It is easy to prove that when and only when $t \in \mathbf{N}^{*}$,
$$k=-\frac{1}{2}(m+1)+\frac{1}{6} t \in \mathbf{N} .$$

Therefore, it is converted to finding all positive integer solutions of the indeterminate equation
$$t^{2}-33 m^{2}=3$$

For this purpose, we first provide all solutions of
$$x^{2}-33 y^{2}=1$$

Direct calculation shows that the fundamental solution of (14) is $\left(x_{0}, y_{0}\right)=(23,4)$, so all positive integer solutions $\left(x_{n}, y_{n}\right)$ of (14) are determined by:
$$x_{n}+y_{n} \sqrt{33}=(23+4 \sqrt{33})^{n} .$$

Notice that, $\left(t_{0}, m_{0}\right)=(6,1)$ is a positive integer solution of (13), therefore, for any positive integer solution $(x, y)$ of (14), the positive integer pair $(t, m)$ satisfying
$$t+m \sqrt{33}=(6+\sqrt{33}) \cdot(x+y \sqrt{33})$$

is a positive integer solution of (13).
In fact, by (15) we know
$$t-m \sqrt{33}=(6-\sqrt{33})(x-y \sqrt{33}),$$

thus
$$\begin{aligned}
t^{2}-33 m^{2} & =(6-\sqrt{33})(6+\sqrt{33})(x-y \sqrt{33})(x+y \sqrt{33}) \\
& =3\left(x^{2}-33 y^{2}\right)=3
\end{aligned}$$

On the other hand, we prove: the positive integer pair $(t, m)$ determined by (15) is all positive integer solutions of (13).

Suppose $(t, m)$ is a positive integer solution of (13), then $3 \mid t^{2}$, hence $3 \mid t$. Combining this with (13), we know $(x, y)=\left(2 t-11 m, \frac{6 m-t}{3}\right)$ (this relationship is obtained by solving (15) in reverse)
is a pair of positive integers, and
$$\begin{aligned}
x^{2}-33 y^{2} & =(2 t-11 m)^{2}-33\left(\frac{6 m-t}{3}\right)^{2} \\
& =\frac{1}{3}\left(3(2 t-11 m)^{2}-11(6 m-t)^{2}\right) \\
& =\frac{1}{3}\left(t^{2}-33 m^{2}\right)=1
\end{aligned}$$

Thus $x, y$ is a positive integer solution of (13), and
$$t=6 x+33 y, m=x+6 y$$

which is consistent with the relationship between $(t, m)$ and $(x, y)$ corresponding to (15).
In summary, the set of all positive integers $m$ that meet the requirement is $\left\{m_{n} \mid m_{n}=x_{n}+6 y_{n}\right.$, where $x_{n}, y_{n} \in \mathbf{N}^{*}$, determined by $x_{n}+y_{n} \sqrt{33}=(23+4 \sqrt{33})^{n}, n \in \mathbf{N}^{*} \}$.