Example 3 In the quadratic equation $x^{2}+z_{1} x+z_{2}+m=0$ about $x$, $z_{1}, z_{2}, m$ are all complex numbers, and $z_{1}^{2}-4 z_{2}=16+$ 20i. Let the two roots of this equation be $\alpha, \beta$. Satisfy $|\alpha-\beta|=2 \sqrt{7}$, find the maximum and minimum values of $|m|$.
Solution 1: By Vieta's formulas, we have
$$
\alpha+\beta=-z_{1}, \alpha \beta=z_{2}+m \text {. }
$$

Since $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta=z_{1}^{2}-4 z_{2}-4 m$, therefore,
$|\alpha-\beta|^{2}=\left|4 m-\left(z_{1}^{2}-4 z_{2}\right)\right|=28$, which means $|m-(4+5 i)|=7$, indicating that the complex number $m$ lies on the circumference of a circle with center $A(4,5)$ and radius 7. Since $|O A|=\sqrt{4^{2}+5^{2}}=\sqrt{41}<7$, the origin $O$ is inside the circle $\odot A$.

Thus, extending the line segment $O A$ to intersect $\odot A$ at points $B$ and $C$, we have $|O B|=|O A|+|A B|=\sqrt{41}+7$, which is the maximum value of $|m|$, and $|O C|=|C A|-|A O|=7-\sqrt{41}$, which is the minimum value of $|m|$.
Therefore, the maximum value of $|m|$ is $7+\sqrt{41}$, and the minimum value of $|m|$ is $7-\sqrt{41}$.
Solution 2: From Solution 1, we have $|m-(4+5 i)|=7$. Let $m=x+i y$, then $\left\{\begin{array}{l}x=7 \cos \alpha+4 \\ y=7 \sin \alpha+5\end{array}\right.$. Therefore, $|m|^{2}=x^{2}+y^{2}=90+56 \cos \alpha+70 \sin \alpha=90+14 \sqrt{41} \sin (\alpha+\psi)$ (where $\sin \psi=\frac{4}{\sqrt{41}}$). Thus, the maximum value of $|m|$ is $\sqrt{90+14 \sqrt{41}}=7+\sqrt{41}$, and the minimum value of $|m|$ is $\sqrt{90-14 \sqrt{41}}=7-\sqrt{41}$.