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?
Actually you have proved that, for any positive integer $x$, $x+1$ is invertible modulo $x^3$.
Indeed, $(x+1)(x^2-x+1)=x^3+1\equiv1\pmod{x^3}$.
No need to check that $x^2-x+1>0$ (which it is, and it may be just remarked). If you find $u\in\mathbb{Z}$ such that $xu\equiv 1\pmod{x^3}$, then you also find $v>0$ such that $xv\equiv1\pmod{x^3}$: just keep adding $x^3$ to $u$ until you get past $0$.
A few more notes.
The $x>1$ clause is just to avoid congruences modulo $1$, which however don't pose relevant problems, do they?
By the way, I'd also state that the restriction for $x$ to be a positive integer can be replaced by “$x$ is an integer”. Even $x=0$ is good, because congruence modulo $0$ is just equality and $0+1=1$ is certainly invertible modulo $0$.
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.
Generally we can invert a unit + nilpotent by a method similar to rationalizing the denominator - see the method of simpler multiples