Example 10 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.
To prove this problem, we need the following lemma.

Lemma: For a real-coefficient polynomial equation of degree $n$, $a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}=0\left(n>1, a_{n} a_{0} \neq 0\right)$, with roots $x_{1}, x_{2}, \cdots, x_{n}$. Let $\Delta_{1}=(n-1) a_{n-1}^{2}-2 n a_{n-2} a_{n}, \Delta_{2}=(n-1) a_{1}^{2}-2 n a_{2} a_{0}$, then
(1) When $x_{1}, x_{2}, \cdots, x_{n}$ are all real numbers, $\Delta_{1} \geqslant 0$ and $\Delta_{2} \geqslant 0$;
(2) When $\Delta_{1}3) .
$$

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

Thus, $t_{2}=\frac{1}{2}\left[\left(2 S_{1}\right)^{2}-2\left(S_{1}^{2}-2 S_{2}\right)-2(n-2)\right]=\frac{1}{12}\left(7 n^{2}-30 n+35\right)$.
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 $x^{n-2}-t_{1} x^{n-1}+t_{2} x^{n-2}+\cdots+t_{n-3} x+t_{n-2}=0$. Using Lemma (1), we have $\Delta_{1}=(n-3)\left(-t_{1}\right)^{2}-2(n-2) t_{2} \geqslant 0$, i.e., $(n-3)(n-1)^{2}-2(n-2) \cdot \frac{7 n^{2}-30 n+35}{12} \geqslant 0$, solving this gives $n \leqslant 5+\sqrt{12}<9$, i.e., $n \leqslant 8$.
When $n=8$, the equation becomes $\left(8 z^{6}+28 z^{5}+56 z^{4}+70 z^{3}+56 z^{2}+28 z+8\right) z=0$, whose non-zero solutions are the 6 roots of the equation $4 z^{6}+14 z^{5}+28 z^{4}+35 z^{3}+28 z^{2}+14 z+4=0 \quad(\star)$.
The equation $(*)$ can be transformed into $4\left(z^{3}+z^{-3}\right)+14\left(z+z^{-2}\right)+28\left(z+z^{-1}\right)+35=0$,
i.e., $4\left(z+z^{-1}\right)^{3}+14\left(z+z^{-1}\right)^{2}+16\left(z+z^{-1}\right)+7=0$.
Since $\Delta_{2}=(3-1) \cdot 16^{2}-2 \cdot 3 \cdot 14 \cdot 7=-76<0$, by Lemma (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 real. 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 problem's conditions.

When $n=7$, the equation can be transformed into $7(z+1) z\left(z^{2}+z+1\right)^{2}=0$. Its non-zero solutions are -1 and $e^{ \pm i \frac{2 z}{3}}$, all of which lie on the unit circle.
In summary, the largest integer $n$ that satisfies the problem's conditions is 7.