Solution:

We show that, if $p$ is a prime number dividing $x_{n}$, then $p$ does not divide $n$.
If $p=2$, then the claim is trivially true, since all terms $x_{n}$ are odd for $n>1$. Hence 2 divides $x_{n}$ if and only if $n=1$.
Suppose then $p>2$. Let $k$ be the smallest positive integer such that $p$ divides $x_{k}$. The sequence $r_{n}$ of the remainders of $x_{n}$ upon division by $p$ takes only finitely many values (between 0 and $p-1$), so there exist two integers $i<j$ such that $r_{i}=r_{j}$. For every $m \geq 0$ we then have $r_{i+m}=r_{j+m}$, so the sequence $r_{n}$ is periodic for $n \geq i$; it follows that the value $r_{k}=0$ ($p$ divides $x_{k}$) must be attained before two terms of the sequence $r_{n}$ repeat. Moreover, if $n<k$, then $r_{n} \neq p-1$, because otherwise $r_{n+1}=r_{n+2}=r_{n+3}=\cdots=1$, and hence $r_{n}$ would never be zero. Therefore we must have $r_{k}=0$ for some $k \leq p-1$. But if $r_{k}=0$, then $r_{k+1}=p-1$ and $r_{n}=1$ for every $n>k+1$, and thus $x_{k}$ is the only term of the sequence divisible by $p$. Since $p$ does not divide any of the numbers $1,2, \ldots, p-1$, $p$ cannot divide $k$.