Example 1 Find all non-square numbers $n \in \mathbf{N}_{+}$, such that $[\sqrt{n}]^{3} \mid n^{2}$. (1)
(3rd Northern Mathematical Olympiad Invitational Competition)
【Analysis】Obviously, $[\sqrt{n}]=1$, i.e., $n=2,3$ satisfy the condition.

Below, let $[\sqrt{n}] \geqslant 2$. Let $A=[\sqrt{n}]$, and $n$ is a non-perfect square. Therefore,
$$
\begin{array}{l}
A<\sqrt{n}<A+1, \\
A^{2}<n<A^{2}+2 A+1 .
\end{array}
$$

Thus, $A^{2}+1 \leqslant n \leqslant A^{2}+2 A$.
Let $n=A^{2}+k(k \in\{1,2, \cdots, 2 A\})$. Then $n^{2}=A^{4}+2 A^{2} k+k^{2}$.
By $A^{3} \mid n^{2}$, we know $A^{2} \mid n^{2}$, so $A \mid k$, i.e., $k=A$ or $2 A$.
Thus, $A^{3} \mid \left(A^{4}+2 A^{2} k\right)$
$$
\Rightarrow A^{3} \mid k^{2} \Rightarrow k=2 A \text {, and } A \mid 4 \text {. }
$$

Therefore, $n=A^{2}+2 A$.
Hence, $A=2, n=8 ; A=4, n=24$.
In summary, $n=2,3,8,24$ satisfy the condition.