Obviously, if $x=1$, then $p=7$.

When $x\ne 1$, then $p$ divides $\frac{x^7-1}{x-1}$. Hence, $x^7\equiv 1\pmod{p}$. This implies $p\not\mid x$. Then, by Fermat's little theorem, we have $x^{p-1}\equiv 1\pmod{p}$. This is followed by $$x^{(7, p-1)}\equiv 1\pmod{p}$$

where $(7, p-1)$ is the greatest common divisor of $7$ and $(p-1)$.

If $p\not\equiv 1\pmod{7}$, i.e. $7\not\mid (p-1)$, then $(7, p-1)=1$. This will lead to $x\equiv 1\pmod{p}$ by the equation above. Then, $$x^6+x^5+\cdots + 1\equiv 1^6+1^6+\cdots+1^6\equiv 7\pmod{p}$$

and $$p\mid(x^6+x^5+\cdots+1)\implies x^6+x^5+\cdots + 1\equiv 0\pmod{p}$$

This will lead to the conclusion $p=7$. Hence, we find the claim holds.