$-1 . D$
Let $x_{0}=\sqrt{x+1}$, the equation can be rewritten as
$$\left|x_{0}-2\right|+\left|x_{0}-3\right|=1 \text {. }$$

Its geometric meaning is: on the number line, the sum of the distances from point $x_{0}$ to points 2 and 3 is 1.

Therefore, $x_{0}$ can take any real number satisfying $2 \leqslant x_{0} \leqslant 3$. Hence, $x=x_{0}^{2}-1$ can take infinitely many values.