Easy to prove by induction that $x_n=2^{2^{n-1}}+2^{-2^{n-1}}+1$. Hence, clearly $p \ne 2$ and hence $p \mid x^2+x+1$ where $x=2^{2^{2020}}$. But all prime divisors of numbers of the form $x^2+x+1$ for any $x \in \mathbb{N}$ are either $p=3$ or $p \equiv 1 \mod 3$. This is classical and easy to prove as follows: Since $p \mid x^3-1=(x-1)(x^2+x+1)$, the order of $x$ modulo $p$ is either $1$ or $3$. If it's $1$, then $x \equiv 1 \mod p$ and hence $p \mid 1^2+1+1=3$ so that $p=3$. On the other hand, if the order is $3$, then by Fermat it divides $p-1$ and hence $p \equiv 1 \mod 3$ as desired.

Tintarn wrote: Easy to prove by induction that $x_n=2^{2^{n-1}}+2^{-2^{n-1}}+1$. Hence, clearly $p \ne 2$ and hence $p \mid x^2+x+1$ where $x=2^{2^{2020}}$. But all prime divisors of numbers of the form $x^2+x+1$ for any $x \in \mathbb{N}$ are either $p=3$ or $p \equiv 1 \mod 3$. This is classical and easy to prove as follows: Since $p \mid x^3-1=(x-1)(x^2+x+1)$, the order of $x$ modulo $p$ is either $1$ or $3$. If it's $1$, then $x \equiv 1 \mod p$ and hence $p \mid 1^2+1+1=3$ so that $p=3$. On the other hand, if the order is $3$, then by Fermat it divides $p-1$ and hence $p \equiv 1 \mod 3$ as desired. Yes, I think this is the shortest and most elegant solution. This problem is also mine. Another way to show that all prime factors of $x^2+x+1$ have this form is to use that the discriminant is a quadratic residue modulo $p$.