Example 4 Let $a, b$ be positive integers such that
$$
\frac{a^{2}+a b+b^{2}}{a b-1}=k\left(k \in \mathbf{Z}_{+}\right) \text {. }
$$

Find the value of $k$.
Let $(a, b)$ be a known positive integer solution of the indeterminate equation, satisfying $a \geqslant b$ and $b$ being the smallest among all positive integer solutions.
Consider the quadratic equation in $x$
$$
x^{2}+b x+b^{2}-k(b x-1)=0 \text {, }
$$

i.e., $x^{2}+(1-k) b x+k+b^{2}=0$
has a positive integer solution $a$, and let the other solution be $a^{\prime}$.
By Vieta's formulas, we know
$$
a^{\prime}=(k-1) b-a \in \mathbf{Z} \text {. }
$$

Also, $a a^{\prime}=b^{2}+k>0$, so $a^{\prime} \in \mathbf{Z}_{+}$.
Since $a^{\prime}$ is a solution to equation (1), $(a^{\prime}, b)$ is also a positive integer solution to the original equation.

By symmetry, $(b, a^{\prime})$ is also a positive integer solution to the original equation.
Combining $a^{\prime} 、 b \in \mathbf{Z}_{+}$ and the minimality of $b$, we get
$$
a^{\prime}=\frac{b^{2}+k}{a} \geqslant b \text {. }
$$

Thus, $k \geqslant a b-b^{2}$
$$
\begin{array}{l}
\Leftrightarrow \frac{a^{2}+a b+b^{2}}{a b-1} \geqslant a b-b^{2} \\
\Leftrightarrow(a b-1) b-\left(a+b+b^{3}\right) \leqslant 0 .
\end{array}
$$
(1) If $a=b$.
$$
\begin{array}{l}
\text { Then } k=\frac{3 a^{2}}{a^{2}-1}=3+\frac{3}{a^{2}-1} \\
\Rightarrow\left(a^{2}-1\right) \mid 3 \Rightarrow a^{2}-1=1 \text { or } 3 \\
\Rightarrow a=2 \Rightarrow b=2, k=4 .
\end{array}
$$
(2) If $a>b$.
(i) When $b \geqslant 4$,
$$
\begin{array}{l}
(a b-1) b-\left(a+b+b^{3}\right) \\
=a\left(b^{2}-1\right)-\left(b^{3}+2 b\right) \\
\geqslant(b+1)\left(b^{2}-1\right)-\left(b^{3}+2 b\right)=b^{2}-3 b-1 \\
=(b-2)(b-1)-3>0 .
\end{array}
$$

This contradicts equation (2).
(ii) When $b=3$,
$$
\begin{array}{l}
(a b-1) b-\left(a+b+b^{3}\right) \\
=3(3 a-1)-(a+3+27)=8 a-33 .
\end{array}
$$

In this case, if $a \geqslant 5$, then $8 a-33>0$, which contradicts equation (2). Therefore, $a=4, k=\frac{37}{11}$, which contradicts $k \in \mathbf{Z}_{+}$.
(iii) When $b=2$, from $k=\frac{a^{2}+2 a+4}{2 a-1} \in \mathbf{Z}_{+}$, we get $4 k=\frac{4 a^{2}+8 a+16}{2 a-1}=2 a+5+\frac{21}{2 a-1} \in \mathbf{Z}_{+}$. Hence, $(2 a-1) \mid 21 \Rightarrow a=1,2,4,11$.
Since $a>b=2$, then $a=4,11 \Rightarrow k=4,7$.
(iv) When $b=1$, $k=\frac{a^{2}+a+1}{a-1}=a+2+\frac{3}{a-1} \in \mathbf{Z}_{+}$.
Hence, $(a-1) \mid 3 \Rightarrow a=2,4 \Rightarrow k=7$.
In summary, the values of $k$ are 4 or 7.