Does $(x+1)^p \equiv x^p+1 \pmod {p^3} \iff x^3 \equiv 1 \pmod p$?

Question

I'm attempting to work out a solution to Are there any solutions to the following congruences where $t^3 \not\equiv 1 \pmod {p^3}$

For $1 < x < p-1$, the following surprising conjecture seems to hold for primes $p<5000$:

$$(x+1)^p \equiv x^p+1 \pmod {p^3} \implies x^3 \equiv 1 \pmod p$$

I think I have a lengthy demonstration of the converse, but haven't found a way to show this direction. Any ideas or suggestions would be great!

Show that $(x+1)^p \equiv x^p +1 \pmod{p^2}$ given conditions on $x$ and $p$. is another very similar problem.