In the quadratic equation $x^{2}+z_{1} x+z_{2}+m=0$, $z_{1}, z_{2}, m$ are all complex numbers, and $z_{1}^{2}-4 z_{2}=16+20 i$. Suppose the two roots $\alpha, \beta$ of this equation satisfy $|\alpha-\beta|=2 \sqrt{7}$, find the maximum and minimum values of $|m|$.
According to Vieta's theorem, we have
$$
\left\{\begin{array}{l}
\alpha+\beta=-z_{1} \\
\alpha \beta=z_{2}+m
\end{array}\right.
$$
$$
\begin{array}{l}
\because(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta=z_{1}^{2}-4 z_{2}-4 m, \\
\therefore \quad|\alpha-\beta|^{2}=\left|4 m-\left(z_{1}^{2}-4 z_{2}\right)\right|=28 \\
\therefore \quad\left|m-\frac{1}{4}\left(z_{1}^{2}-4 z_{2}\right)\right|=7
\end{array}
$$

This implies
$$
|m-4+5 i|=7
$$

This indicates that the complex number $m$ lies on the circumference of a circle with center $A(4,5)$ and radius 7. Since $\because|O A|=\sqrt{4^{2}+5^{2}}=\sqrt{41}<7$, the origin $O$ is inside the circle $\odot A$. Connecting $O A$ and extending it to intersect $\odot A$ at points $B$ and $C$, we have
$|O B|=|O A|+|A B|=\sqrt{41}+7$ as the maximum value of $|m|$,
$|O C|=|C A|-|A O|=7-\sqrt{41}$ as the minimum value of $|m|$.
$\therefore \quad|m|$ has a maximum value of $\sqrt{41}+7$,
and a minimum value of $7-\sqrt{41}$.