9. (16 points) Given $x, y \in \mathbf{R}$, and satisfying $(x+1)(y-1)=1$.
Find the minimum value of $\left(x^{2}+1\right)\left(y^{2}+1\right)$.
$$
\begin{array}{l}
(x+1)(y-1)=1 \\
\Rightarrow x y-x+y-1=1 \\
\Rightarrow(x y+1)+(y-x)=3 . \\
\text { And }\left(x^{2}+1\right)\left(y^{2}+1\right) \\
=(x y+1)^{2}+(y-x)^{2} \\
\geqslant \frac{1}{2}[(x y+1)+(y-x)]^{2}=\frac{9}{2} .
\end{array}
$$

When and only when $x y+1=y-x=\frac{3}{2}$, that is,
$$
x=\frac{\sqrt{17}-3}{4}, y=\frac{\sqrt{17}+3}{4}
$$

or $x=-\frac{\sqrt{17}+3}{4}, y=\frac{3-\sqrt{17}}{4}$
the equality holds.
Therefore, the minimum value sought is $\frac{9}{2}$.