Let $ n = 2^c - 1$, and $ a = kc + r,\ r < c$, then $ 2^a\equiv (2^c)^k\cdot 2^r\equiv 2^r\pmod n$. If $ r_a = r_b = c - 1$, then $ 2^a + 2^b + 1\equiv 2\mod 2^c - 1$. If $ c > 3$ and $ r_a < c - 1$ or $ r_b < c - 1$, then $ 2^{r_a} + 2^{r_b} + 1 < 2^c - 1$. Therefore for all solutions $ c\le 3$. 1. $ c = 1$ a,b - any. 2. $ c = 2$ a,b are even. 3. $ c = 3$ $ a = 1\mod 3,b = 2\mod 3$ or $ a = 2\mod 3,b = 1\mod 3$. ModEdit: correct but sloppy, needs rewrite.