If $p=2$, then it is easy, as $x_1+x_2 \equiv 0 \pmod{2} \iff x_1 \equiv x_2 \pmod{2}$. Assume $p>2$. Lemma: If $x_1,x_2,…,x_p$ satisfies $p|\sum_{i=1}^{p} x_i^{p-1}$, then $p|\sum_{i=1}^{p} (x_i-a)^{p-1}$, where $a$ is an integer. As $(x_i-a)^{p-1} = x_i^{p-1}-a\binom {p-1}{2} x_i^{p-2}+\dots+a^{p-1}$, we have that $\sum_{i=1}^{p} (x_i-a)^{p-1} = \sum_{i=1}^{p} x_i^{p-1} - a\binom {p-1}{2}\sum_{i=1}^{p} x_i^{p-2} + a^2\binom {p-1}{3}\sum_{i=1}^{p} x_i^{p-3} - \dots + p \cdot a^{p-1} \equiv 0 + 0 + 0 + \dots + 0 \equiv 0 \pmod{p}$ Take $a = x_1$. Then, by the lemma $\sum_{i=2}^{p} (x_i-x_1)^{p-1} \equiv 0 \pmod{p}$ Due to Fermat’s Little Theorem, these terms are either $1$’s or $0$’s. However, there are only $p-1$ of these terms. Therefore, none of the terms can be a $1$; otherwise, you would have to add at most $p-1$ ones to get something at least $p$, contradiction. Therefore, all of these terms must be zeroes, which happens iff $(p,x_i-x_1) = p$ for all $i: 1 \leq i \leq p$. Therefore, $x_2 \equiv x_3 \equiv \dots \equiv x_p \equiv x_1 \pmod{p}$

If we let $s_k=\displaystyle\sum\limits_{1\le a_1<a_2<...<a_k\le p} x_{a_1} x_{a_2}... x_{a_k}$ for $0\le k<p$ (with $s_0=1$) , then $$s_k=\dfrac{1}{k}\displaystyle\sum\limits_{i=1}^{k} (-1)^{i+1} (x_1^i+x_2^i+...+x_p^i)\cdot s_{k-i} $$so $s_k$ is divisible by $p$. Hence, $$(x-a_1)(x-a_2)...(x-a_p)=x^p-s_1x^{p-1}+...+s_{p-1}x-a_1...a_p\equiv x^p-a_1a_2...a_p (\mathrm{mod}\ p)$$ Plugging in $a_i$ in the previous relation and using FLT, we get that $a_i\equiv a_1....a_p (\mathrm{mod}\ p)$ for any $i$, whence the conclusion.

From Newton's identities, it directly follows that each symmetric polynomial $e_{i}$ $1\leq i\leq p -1$ is divisible by $p$. This in turn implies that all $x_i$ are roots of a polynomial $x^p + \alpha$ over $\mathbb{F}_p$. Fermat's little theorem now gives $x_1=x_2=\dots x_p=-\alpha \pmod{p}$.