$\frac{\sqrt{x}+1}{\sqrt{x}+3}\in \mathbb{Z}$
$\Rightarrow \sqrt{x}+1\vdots \sqrt{x}+3$
$\Rightarrow \sqrt{x}+3-2\vdots \sqrt{x}+3$
$\Rightarrow -2\vdots \sqrt{x}+3$
$\Rightarrow \sqrt{x}+3\in Ư(-2)=\{ \pm 1; \pm 2\}$
$\sqrt{x}+3\ge 3$
$\Rightarrow$ Không có x thoả mãn.