Let $ T_p$ is half period, suth that $ p|2^p + 1$ and $ p|2^n + 1\to n = mT_p,m = 1\mod 2$. If $ n|2^n + 2$ then for all odd prime divisor $ p|n$ we get $ \frac {n - 1}{T_p}$ is odd integer. Therefore $ v_2(T_p)$ is same for all odd prime divisors. If $ n = p$ (is prime) we get $ p|4\to p = 2$. Obviosly $ v_2(n)\le 1$. If $ n = 2p$, then $ p|2^2 + 2 = 6\to n = 6$. From conditions we get, that n had odd prime divisor $ p\le 43$ and $ T_3 = 1,T_5 = 2$, $ T_7$ is not defined, because period is 3 and $ 7\not |2^k + 1$. $ T_{11} = 5, T_{13} = 6,T_{17} = 4, T_{19} = 9,T_{23} = 11,T_{29} = 14,$ $ T_{31}$ is not defined. $ T_{37} = 18,T_{41} = 10,T_{43} = 7$. If $ p|n$ and $ T_p$ is even, then $ p = 1\mod 4$, therefore $ n - 1$ must be even and $ n = pk, \frac {k - 1}{T_p} - \ odd$. It mean, that $ v_2(k - 1) = v_2(T_p) < v_2(p - 1)$, therefore k had prime divisor q, suth that $ v_2(q - 1)\le v_2(T_p)$ and $ v_2(T_q) < v_2(T_p)$ - contradition. We get, that all odd prime divisors n had odd $ T_p$, therefore n=2m, m- odd. Therefore one one prime divisor of n $ \le 31$. It is $ p = 3$ or $ p = 19$ or $ p = 23$. If $ n = 2*23*k$, then $ 11|2k - 1$. But $ k = 17,39$ not work. If $ n = 2*19k$, then $ 9|2k - 1, k = 5,23,41,59$ not work. Therefore $ n = 2*3^l*k$. Obviosly $ 2^n\equiv 1\mod 3^{l+1}$, therefore $ l$ can not be $ >1$. Therefore $ n=6k$, were $ k$ odd prime $ > 43$ or $ k = 1$, because prime divisors of k must be $ >43$ If k is prime $ k|2^6+2$. If $ k$ is odd prime $ T_k|6-1 = 5\to k=11$. Therefore all solutions $ n|2^n + 2,n\le 1997$ are $ n = 1,2,6,66.$

Rust wrote: Let $ T_p$ is half period, suth that $ p|2^p + 1$ and $ p|2^n + 1\to n = mT_p,m = 1\mod 2$. Heh? And it's wrong anyhow, 946 is a solution.

http://www.kalva.demon.co.uk/apmo/asoln/asol972.html

Related sequence: http://oeis.org/A006517