Problem on solving congruence equation

Question

I want to find $x$ in this equation for known positive integer values $a$, $b$ and $m$. $$a-bx+2^{6x+1}\equiv0\ \pmod{m}$$ Is the number of solutions for $x$ finite? Any help will be appreciated.

Answer 1

If $u$ is a solution, then $u+km\phi(m)$ is also a solution , if m is odd.

Proof : $$a-b(u+km\phi(m))+2^{6(u+km\phi(m))+1} \equiv a-bu+2^{6u+1}*2^{6km\phi(m)} \equiv a-bu+2^{6u+1}\equiv 0\ mod\ (\ m\ )$$ for odd m

For even $m$, the equation $2^{\phi(m)} \equiv 1\ (\ mod\ m\ )$ does not hold and carmichaels function might help.

So, if there is a solution, there are infinite many, if m is odd.