$x+1$ is always invertible in $\Bbb Z_{x^3}$. 
Claim: For any $x > 1$ and $x \in \Bbb N$, $x+1$ is always invertible in $\Bbb Z_{x^3}$.

Proof: We know that $x^3 +1 = (x+1)(x^2-x+1)$. Since $x > 1$ and $x \in \Bbb N$, we have $x^2-x+1 >0$ and $(x+1)(x^2-x+1) = x^3 +1 \equiv 1 \mod x^3$. Hence $x+1$ is always invertible in $\Bbb Z_{x^3}$.
Is the proof correct?

You proved the Bezout equation $\,1 = (\color{#c00}{x+1})f(x) - \color{#c00}{x^3}\,$ for $\,f(x) \in \Bbb Z[x]\,$
This implies that $\,\color{#c00}{x+1}\,$ and $\,\color{#c00}{x^3}\,$ are coprime in every ring, since any common divisor divides the RHS so also the LHS $= 1$, i.e. every common divisor is a unit (invertible).
$\!\bmod x^3\!:\ $ reducing Bezout $\, \Rightarrow 1\equiv (x+1)f(x)\,$ $\Rightarrow x+1\,$ is invertible
