Problem 4

Putting $m=3$ into $1+m=2^{x-2}(m^2-8)$ gives $4=2^{x-2}$, hence $x=4$. Then $y=2^{x-1}m-1=23$, and the symmetry $y\mapsto-y$ gives $(4,\pm23)$. Together with Step 2, the complete set is $(0,\pm2)$ and $(4,\pm23)$.