Example 45 Let $a, b, c, k$ be positive integers, $\frac{(a+b+c)^{2}}{a b c}=k$, find all possible values of $k$.

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Let's assume for some $k$, $a+b+c$ is minimal. Let $a \leqslant b \leqslant c$. The equation in terms of $c$, $c^{2}-(k a b-2 a-2 b) c+(a+b)^{2}=0$, has a root $c'$. By Vieta's formulas, $c'$ is a positive integer, thus
$c^{2} \leqslant c' c=(a+b)^{2}$, hence $c \leqslant a+b$. It is easy to verify that

$$
\begin{aligned}
k=f(c)= & \frac{(a+b+c)^{2}}{a b c}=\frac{2(a+b)}{a b}+\frac{c}{a b}+\frac{(a+b)^{2}}{a b c} \leqslant f(b) \\
& =\frac{2(a+b)}{a b}+\frac{1}{a}+\frac{(a+b)^{2}}{a b^{2}}=\frac{4}{a}+\frac{4}{b}+\frac{a}{b^{2}} \\
& \leqslant \frac{4}{a}+\frac{5}{b} \leqslant 9, \text{ so } k \in\{1,2, \cdots, 9\} .
\end{aligned}
$$

Let $a=b=c=x$, then $k=\frac{9}{x}$, $k$ can be $1,3,9$.
Let $a=x, b=2 x, c=3 x, k=\frac{6}{x}$, $k$ can be 2,6.
Let $a=1, b=4, c=5, k=5$.
Let $a=x, b=x, c=2 x, k=\frac{8}{x}$, $k$ can be 4,8.
Next, we prove that $k \neq 7$.
If $k=7$, then $7 \leqslant \frac{4}{a}+\frac{5}{b}$, $a$ must be 1, so $b$ must also be 1. Hence $7=\frac{(2+c)^{2}}{c}=4+c+\frac{4}{c}$, it is easy to see that $c+\frac{4}{c} \geqslant 2 \sqrt{4}=4>3$, so $k=1,2,3,4,5,6,8,9$.