4. Given real numbers $x, y$ satisfy $x y+1=4 x+y$, and $x>1$. Then the minimum value of $(x+1)(y+2)$ is $\qquad$

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4.27.

From the problem, we know $y=\frac{4 x-1}{x-1}$.
$$
\begin{array}{l} 
\text { Therefore, }(x+1)(y+2)=(x+1)\left(\frac{4 x-1}{x-1}+2\right) \\
\quad= \frac{3(x+1)(2 x-1)}{x-1} .
\end{array}
$$

Let $x-1=t>0$. Then
$$
\begin{array}{l}
(x+1)(y+2)=\frac{3(t+2)(2 t+1)}{t} \\
=6\left(t+\frac{1}{t}\right)+15 \geqslant 27,
\end{array}
$$

when and only when $t=\frac{1}{t}$, i.e., $t=1, x=2, y=7$, the equality holds.
Thus, the minimum value of $(x+1)(y+2)$ is 27.