Is this corect ? Put $n=p-1$ to deduce all of them are not divisible by $p$ Then denote $S_k$ denote the k-th symmetric sum and $P_k=x_1^k+x_2^k+\cdots+x_p^k$ We have in $\mathbb F_p$ that $P_1=P_2=\cdots=P_{p-1}=0$ and by newton sums we have $S_1=S_2=\cdots=S_{p-1}=0$ and we consider the polynomial $P(X)=(X-x_1)(X-x_2)\cdots (X-x_p)$ which in $F_p$ is $X^p-x_1x_2\dots x_p$ and the roots are $x_1,\cdots x_p$ so we get $x_1^p=x_2^p=x_1x_2\cdots x_p$ but $x^p=x$ and we are done

Honestly i should post NT more often, this one is nice though. If $p=2$ we get $2 \mid x_1+x_2$ so as $1 \equiv -1 \pmod 2$ we have $2 \mid x_1-x_2$ as desired. Now if $p \ge 3$, consider the polynomial $P(x)=(x-x_1)(x-x_2)...(x-x_p)$, clearly $\text{deg} P=p$ and $P \in \mathbb Z[x]$. And therefore we can let $P(x)=x^p+a_{p-1}x^{p-1}+...+a_1x+a_0$ where all $a_j$'s are integers, so by taking newton sums modulo $p$ we have $p \mid a_{p-1},...,a_1$ which gives that $P(x) \equiv x^p+a_0 \equiv x+a_0 \pmod p$ therefore by putting $x_i$ in this we get that $x_i \equiv -a_0 \pmod p$ for any $1 \le i \le p$ which proves that all $x_i$'s are congrent modulo $p$ and therefore we are done .

Let us work in $\mathbb{F}_p$. Suppose $r$ of the $x_i$'s are nonzero and WLOG they are $x_1,\ldots,x_r$. Let $g$ be a generator of $\mathbb{F}_p^\times$. Let $x_i=:g^{\alpha_i}$ where $0\leq\alpha_i\leq p-2$ is an integer for $i=1,\ldots,r$. Then the condition becomes $(g^n)^{\alpha_1}+\cdots+(g^n)^{\alpha_r}=0$ for all $n\in\mathbb{Z}$. It follows that the elements of $\mathbb{F}_p^\times$ are roots of $P(x):=x^{\alpha_1}+\cdots+x^{\alpha_r}\in\mathbb{F}_p[x]$. Thus $P(x)$ is divisible by \[ \prod_{a\in\mathbb{F}_p^\times}(x-a)=x^{p-1}-1. \]Since $\deg P\leq p-2$ if $P$ is nonzero, $P(x)=0$ and thus either $r=0$ in which case we are done, or $r=p$ and all the $\alpha_i$'s are equal, as desired. $\square$