The integers $n$ such that $n | 2^n + 1$ are sequence A006521 in the OEIS: http://oeis.org/A006521
Note the comments. So for example 19 has $2012$ hands in $n$'s hair where $n = 3^2 \times 19^{2012}$.
EDIT: If $p$ is a prime dividing $2^{3^k}+1$ for some integer $k$, then $3^k p$ is in the sequence A006521. For convenience let $f(j) = 2^{2 \cdot 3^j} - 2^{3^j} + 1$. Now $2^{3^k}+1 = 3 \prod_{j=0}^{k-1} f(j)$ Any prime $p > 3$ dividing $f(j)$ divides $2^{3^{j+1}}+1$ but not $2^{3^j}+1$, and thus not $f(k)$ for any $k < j$. Now $2^{3^k} \equiv -1 \mod 9$ for all $k \ge 1$, so $f(k) \equiv 3 \mod 9$ has only one factor of $3$. Thus each $f(j)$ for $j \ge 1$ must be divisible by some prime $p > 3$, which doesn't divide any other $f(k)$. We conclude that there are infinitely many different primes that divide members of A006521. In particular for any $N$, there is a member of A006521 with $N$ distinct prime factors.