Let $f(n)=n^{2}+n+1$. Note that

$$f\left(n^{2}\right)=n^{4}+n^{2}+1=\left(n^{2}+n+1\right)\left(n^{2}-n+1\right)$$

This means that $f(n) \mid f\left(n^{2}\right)$ for every positive integer $n$. By induction on $k$, one can easily see that $f(n) \mid f\left(n^{2^{k}}\right)$ for every positive integers $n$ and $k$. Note that the required condition is equivalent to $f(n) \mid f\left(2^{n}\right)$. From the discussion above, if there exists a positive integer $n$ so that $2^{n}$ can be written as $n^{2^{k}}$, for some positive integer $k$, then $f(n) \mid f\left(2^{n}\right)$. If we choose $n=2^{2^{m}}$ and $k=2^{m}-m$ for some positive integer $m$, then $2^{n}=n^{2^{k}}$ and since there are infinitely many positive integers of the form $n=2^{2^{m}}$, we have the desired result.