Example 8 (2004 National High School Competition Question) Let the function $f: \mathbf{R} \rightarrow \mathbf{R}$, satisfy $f(0)=1$, and for any $x, y \in$ $\mathbf{R}$, there is $f(x y+1)=f(x) \cdot f(y)-f(y)-x+2$, then $f(x)=$
Solution: Fill in $x+1$. Reason: For any $x, y \in \mathbf{R}$,
we have
$$
\begin{array}{l}
f(x y+1)=f(x) \cdot f(y)-f(y)-x+2, \\
f(x y+1)=f(y) \cdot f(x)-f(x)-y+2,
\end{array}
$$

Thus, $f(x) \cdot f(y)-f(y)-x+2=f(y) \cdot f(x)-f(x)-y+2$,
which simplifies to $f(x)+y=f(y)+x$.
By the problem statement, when $y=0$, we have $f(0)=1$, hence $f(x)=x+1$.