Solution:
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.