9. Proof: From $x^{n}+1=y^{n+1}$, we have
$$x^{n}=y^{n+1}-1=(y-1)\left(y^{n}+y^{n-1}+\cdots+1\right),$$

Let $p \mid(y-1)$, then it must be that $p \mid x$. And by $(x, n+1)=1$, we have $p \nmid(n+1)$. Thus, $(y-1, n+1)=1$. We also have
$$y^{n}+y^{n-1}+\cdots+1 \equiv n+1(\bmod y-1),$$

Therefore, $y^{n}+y^{n-1}+\cdots+1$ must be coprime with $y-1$, otherwise, if there exists a prime $p \mid\left(y-1, y^{n}+\cdots+1\right)$, then by (11) we have $p \mid(n+1)$, which contradicts the fact that $y-1$ and $n+1$ are coprime. Thus, there must be $x=x_{1} x_{2},\left(x_{1}, x_{2}\right)$ $=1$, such that
$$x_{1}^{n}=y-1, x_{2}^{n}=y^{n}+y^{n-1}+\cdots+1$$

However, we have
$$y^{n}<1+y+\cdots+y^{n}<(y+1)^{n},$$

Therefore, $1+y+\cdots+y^{n}$ cannot be expressed as the $n$-th power of an integer, which contradicts the second equation in (12).