Example 3 Find the largest integer $n$, such that all non-zero solutions of the equation $(z+1)^{n}=z^{n}+1$ lie on the unit circle.
$$
\begin{array}{l}
\text { Solution: Using the binomial theorem, the equation can be transformed into } \\
z\left(C_{n}^{1} z^{n-2}+C_{n}^{2} z^{n-3}+C_{n}^{3} z^{n-4}+\cdots+C_{n}^{n-2} z+\right. \\
\left.C_{n}^{n-1}\right)=0(n>3) .
\end{array}
$$

Let the non-zero solutions of the equation be $z_{i}(i=1,2, \cdots, n-2)$. By Vieta's formulas, we have
$$
\begin{aligned}
S_{1} & =\sum_{i=1}^{n-2} z_{i}=-\frac{C_{n}^{2}}{C_{n}^{1}}=-\frac{n-1}{2}, \\
S_{2} & =\sum_{1 \leqslant i4$ satisfies the given conditions, then $z_{i} \cdot \bar{z}_{i}=\left|z_{i}\right|^{2}$ $=1, x_{i}=z_{i}+\bar{z}_{i}$ is a real number $(i=1,2, \cdots, n-2)$. Since the coefficients of the equation are all real, the roots appear in conjugate pairs, so the non-zero solutions can also be represented as $\bar{z} i=1,2, \cdots, n-2)$. Therefore,
$$
\begin{array}{l}
t_{1}=\sum_{i=1}^{n-2} x_{i}=\sum_{i=1}^{n-2}\left(z_{i}+\bar{z}_{i}\right)=2 S_{1}=1-n, \\
t_{2}=\sum_{1: i<j \leqslant n-i} x_{i} x_{i}:=\frac{1}{2}\left(t_{1}^{2}-\sum_{i=1}^{n-2} x_{i}^{2}\right), \\
\sum_{i=2}^{n-2} x_{i}^{2}=\sum_{i=1}^{n-2}\left(z_{i}^{2}+\bar{z}_{i}^{2}+2 z_{i} \cdot \bar{z}_{i}\right) \\
=\sum_{i=1}^{n-2}\left(z_{i}^{2}+\bar{z}_{i}^{2}\right)+2 \sum_{i=1}^{n-2} z_{i} \cdot \bar{z}_{i} \\
=2 \sum_{i=1}^{n-2} z_{i}^{2}+2 \sum_{i=1}^{n-2} 1 \\
=2\left(S_{1}^{2}-2 S_{2}\right)+2(n-2) .
\end{array}
$$

Therefore, $t_{2}=\frac{1}{2}\left[\left(2 S_{1}\right)^{2}-2\left(S_{1}^{2}-2 S_{2}\right)\right.$
$$
\begin{aligned}
& -2(n-2)] \\
= & \frac{1}{12}\left(7 n^{2}-30 n+35\right) .
\end{aligned}
$$

By Vieta's formulas, the real numbers $x_{i}(i=1,2, \cdots, n-2)$ are the $n-2$ roots of the real-coefficient equation $x^{n-2}-t_{1} x^{n-3}+t_{2} x^{n-4}+\cdots+$ $t_{n-3} x+t_{n-2}=0$. Using Theorem (1), we have
$$
\Delta_{1}=(n-3)\left(-t_{1}\right)^{2}-2(n-2) t_{2} \geqslant 0,
$$

which is $(n-3)(n-1)^{2}-2(n-2) \cdot$
$$
\frac{7 n^{2}-30 n+35}{12} \geqslant 0 \text {. }
$$

Simplifying, we get
$$
(n-4)\left[(n-5)^{2}-12\right] \leqslant 0 .
$$

Solving, we get $n \leqslant 5+\sqrt{12}<9$, i.e., $n \leqslant 8$.
When $n=8$, the equation becomes
$$
\begin{array}{l}
\left(8 z^{6}+28 z^{5}+56 z^{4}+70 z^{3}+56 z^{2}\right. \\
+28 z+8) z=0 .
\end{array}
$$

Its non-zero solutions are the 6 roots of the following equation (*):
$$
\begin{array}{l}
4 z^{6}+14 z^{5}+28 z^{4}+35 z^{3}+28 z^{2}+14 z+4 \\
=0 . \\
\text { The equation }(*) \text { can be transformed into } \\
4\left(z^{3}+z^{-3}\right)+14\left(z^{2}+z^{-2}\right)+28\left(z+z^{-1}\right) \\
+35=0, \\
4\left(z+z^{-1}\right)^{3}+14\left(z+z^{-1}\right)^{2}+16\left(z+z^{-1}\right) \\
+7=0 .
\end{array}
$$

The equation (*) can be transformed into

$4\left(z+z^{-1}\right)^{3}+14\left(z+z^{-1}\right)^{2}+16\left(z+z^{-1}\right)$

$\Delta_{2}=(3-1) \cdot 16^{2}-2 \cdot 3 \cdot 14 \cdot 7=$ $-76<0$, by Theorem (2), the roots of the equation $4 x^{3}+14 x^{2}+$ $16 x+7=0$ are not all real, i.e., there exists a root $z_{i}$ of equation (*) such that $z_{i}+z_{i}^{-1}$ is not a real number. Thus, $z_{i}+z_{i}^{-1} \neq z_{i}+\bar{z}_{i}$, i.e., $\left|z_{i}\right| \neq 1$. Therefore, $n=8$ does not satisfy the given conditions.
When $n=7$, the equation becomes
$$
\begin{aligned}
0 & =(z+1)^{7}-\left(z^{7}+1\right) \\
& =(z+1) \sum_{i=0}^{6} C_{6}^{i} z^{i}-(z+1) \sum_{i=0}^{6}(-1)^{i} z^{6} \\
& =(z+1) \sum_{i=0}^{6}\left[C_{6}^{i}-(-1)^{i}\right] z^{i} \\
& =7(z+1) \approx\left(z^{2}+z+1\right)^{2} .
\end{aligned}
$$

Its non-zero solutions are -1 and $\cos 120^{\circ} \pm i \sin 120^{\circ}$, all on the unit circle.
In summary, the largest integer $n$ that satisfies the given conditions is
$$
\boxed{7}
$$