If $|b|\leq 2$, evidently no solutions exist. If $|b|>2$, let $r_1<r_2$ be roots of $f$, i.e. $f(x)=(x-r_1)(x-r_2).$ $f(f(x)+x)=(x-r_1)(x-r_1+1)(x-r_2)(x-r_2+1)$ If $|b|\geq\sqrt{5}$, we have $x_2-x_1>1$, $x_2>x_2-1>x_1>x_1-1$. Hence, $x\in (x_1-1,x_1)\cup (x_2-1,x_2)$. Since $|b|>2$, $\exists a\neq 0,\pm 1:\ b=a+\frac{1}{a}\implies x\in (a-1,a)\cup \left(\frac{1}{a}-1,\frac{1}{a}\right)$ If $a\in \mathbb{Z}$ or $\frac{1}{a}\in \mathbb{Z}$, there is a unique solution. Otherwise, there are exactly two solutions. Let $|b|\in (2,\sqrt{5})$, consider two cases, $a<0;\ x\in \lbrace 0,1\rbrace\quad \text{and}\quad a>0;\ x\in \lbrace -1,-2\rbrace$ Therefore, $\text{S1}:\ |b|\leq 2\Rightarrow \text{no solutions}$ $\text{S2}:\ b=a+\frac{1}{a},\ a\in\mathbb{Z},\ a\neq 0,\pm 1\Rightarrow \text{one solution}$ $\text{S3}:\ \text{Otherwise, exactly two solutions}$

I have a solution: Lemma 1: $f(f(x)+x)=f(x)f(x+1)$ So we have $f(x)f(x+1)<0$ so $f(x)<0$ or $f(x+1)<0$ If $f(x)<0$, $x_1<b^2-4<x_2$ ( $x_1$,$x_2$ are root of $f(x)$. And......

Obviously we represent $f(x)=(x+r_1)(x+r_2)$. By Vieta's relations we have that $r_1r_2=1$ and that $r_1+r_2=b$. Thus we easily have that $f(f(x)+x)=(x+r_1)(x+r_2)(x+1+r_1)(x+1+r_2) = f(x)f(x+1)$ Meaning that we have reduced our question to $f(x)f(x+1) < 0$. Meaning that we have 2 cases for examination. The first case is when $f(x) < 0$ but $f(x+1) > 0$ By solving the quadratic inequalities we get that $x \in \left(-1+\frac{-b-\sqrt{b^2-4}}{2},\frac{-b-\sqrt{b^2-4}}{2} \right) \cup \left(-1+\frac{-b+\sqrt{b^2-4}}{2},\frac{-b+\sqrt{b^2-4}}{2} \right)$. Obviously each of the intervals has one integer in himself. Thus we have only $2$ solutions in this case. The second case is when $f(x) > 0$ but $f(x+1) < 0$. By solving the quadratic inequalities we get an empty set. So this implies that we have no solutions in this case.