While we can do it algebraically by factorizing some $2^b - 1$ out of the expression, how do we show it's composite in another way?

One solution that is known is $2^m - 1= (2^b - 1)(2^{m-b} + 2^{m-2b}...+2^{m-ab})$ which is composite. However, does this relation hold when m is not composite? I have always assumed it would, since it's similar to factorizing $x-1$ out of $x^3 - 1$.