I didn't find any solutions for the following Diophantine equation:
for $n=2k-1$, and $a,b\in \{2,3,4,...\}$, and $m\in \mathbb{N}$ $$if\ \ 2^n-1=a\cdot b,\ \ then\ \ a+b\neq 2^m$$
I would like to prove that there aren't any solutions, but I have no idea how.
Because we assume that $a,b>1$ it follows that $2^n=1+ab>a+b=2^m$. Hence $n>m$, and $1<a<2^m-1$. Using $b=2^m-a$ gives $$ 2^n=1+ab=1+a(2^m-a)=1-a^2+2^ma. $$ Reducing this modulo $2^m$ gives $$ a^2-1\equiv0\pmod{2^m}. $$ It is well known that the solutions of $x^2\equiv1\pmod{2^m}$ are the residue classes of $1,2^m-1$ and $2^{m-1}\pm1$ modulo $2^m$. The first two were excluded, so we must have $\{a,b\}=\{2^{m-1}\pm1\}$.
But then $$ 2^n=1+ab=1+(2^{m-1}-1)(2^{m-1}+1)=1+2^{2(m-1)}-1=2^{2(m-1)} $$ violating the assumption that $n$ should be odd.