Let $y=x+1$, then $x=y-1$ and $$x^{2017}=(y-1)^{2017} = (\cdots) + 2017 y - 1$$ This means that the remainder of $(y-1)^{2017}$ being divided by $y^2$ is $(2017y-1)$. Setting $y=x+1$ back we find the answer is $\boxed{2017x+2016}$