Let $x=2^{p-1}-1$ be composite, $p$ prime and $3\mid x$. Why $p \mid x$?

Question

Let $x=2^{p-1}-1$ be a composite odd natural number (a "wrong" Mersenne) and $p$ is prime and $3\mid x$. Why $p \mid x$? Does really $p$ always divides $x$?

Note: We know that $x$ is a sum of powers of two.

Examples:

$1+2+4+8=2^4-1=2^{5-1}-1=15$

$1+2+\ldots+2^{29}=2^{30}-1=2^{31-1}-1=1.073.741.823$

Answer 1

if $p=2$ then $x=1$, but $3|x$, so $p$ is odd prime number.

Based on Fermat's little theorem $2^{p-1}\equiv 1 \pmod{p}$, as $gcd(p,2)=1$.