This is an old competition problem I am struggling with.
How can I prove that the equality $2^m = 4096c^3 + 192c^2+3c+1$ is not valid for any natural numbers $c$ and $m$ (and for $m=0$).
This equation has a solution only if $c=0$ (this is what I am trying to prove).
The discriminant of the cubic equation is $$\Delta=-110592 \,\left(2^{m+6}-63\right)^2 \quad < 0\quad \forall m$$ For the depressed equation $t^3+pt+q=0$, we have $$\color{red}{\large p=0} \qquad\text{and}\qquad \color{red}{\large q=\frac{63-2^{m+6}}{2^{18}}}$$
So, in the real domain, the only solution is $$c=\frac{\Big[2^{m+6}-63\Big]^{\frac 13}-1}{64} $$