Example 3 Find all positive integers $k$ such that the indeterminate equation $x^{2}+y^{2}=k x y-1$ has positive integer solutions for $x$ and $y$.

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Prove that among the positive integer solutions of the given indeterminate equation, the one that minimizes $x+y$ is $\left(x_{0}, y_{0}\right)$.
If $x_{0}=y_{0}$, then $x_{0}^{2} \backslash\left(2 x_{0}^{2}+1\right)$.
Therefore, $x_{0}^{2} \mid 1 \Rightarrow x_{0}=1$.
Thus, $k=\frac{x_{0}^{2}+y_{0}^{2}+1}{x_{0} y_{0}}=3$.
If $x_{0} \neq y_{0}$, by symmetry, assume $x_{0}>y_{0}$.
By $k=\frac{x_{0}^{2}+y_{0}^{2}+1}{x_{0} y_{0}}$
$$
\Rightarrow x_{0}^{2}-t y_{0} x_{0}+y_{0}^{2}+1=0 \text {. }
$$

Viewing equation (1) as a quadratic equation in $x_{0}$, by Vieta's formulas, it has two roots $x_{0}$ and $t y_{0}-x_{0}$, and $t y_{0}-x_{0}=\frac{y_{0}^{2}+1}{x_{0}}$.

Since $t y_{0}-x_{0} \in \mathbf{Z}_{+}$, $\left(\frac{y_{0}^{2}+1}{x_{0}}, y_{0}\right)$ is also a positive integer solution of equation (1).
By $x_{0}>y_{0} \geqslant 1$, we have $\frac{y_{0}^{2}+1}{x_{0}}<x_{0}$, i.e.,
$$
\frac{y_{0}^{2}+1}{x_{0}}+y_{0}<x_{0}+y_{0} \text {. }
$$

This contradicts the minimality of $\left(x_{0}, y_{0}\right)$.
In conclusion, $k=\frac{x_{0}^{2}+y_{0}^{2}+1}{x_{0} y_{0}}=3$.