Is there any efficient approach to answer this question and similar ones?
for $m=0,1,2,3,4,5$ it's easy to see there are no solutions, because the left hand side is at least $32\cdot 1 \cdot 1 + 10 \cdot 1 - 3 \cdot 1 - 1 = 38$. but what about the rest? is there some sophisticated approach to take care of the cases for which $2^m$ is larger than the lower bound mentioned above?
Thanks
There are plenty of solutions : The solutions upto $m=60$ are :
For $m=86$ and $m=98$ , many pairs $(x/y)$ do the job.
Probably for infinite many $m$ , there are solutions (Seems that $m$ must be of the form $4k+2$), but I think it is very difficult to find all such $m's$.