Prove that $2^n +1$ in never a perfect cube

Prove that $2^n +1$ in never a perfect cube

I've been thinking about this problem, but I don't know how to do it. I know that if $m^3=2^n+1$, then $m$ should be an odd number, but I 'm not able to get to a contradiction.

Solution 1:

Suppose that $2^n+1=m^3$. Then $$2^n=m^3-1=(m-1)(m^2+m+1),$$ so each of $m-1$ and $m^2+m+1$ is a power of $2$.

But $m^2+m+1$ is odd, and therefore $m^2+m+1=1$. That forces $m^2+m=0$, giving $m=0$ (impossible) or $m=-1$ (also impossible).

Solution 2:

Andre's answer is neat, I've got a similar solution. Assume $m^3 = 2^n + 1$ then $m$ is odd therefore $m=2p+1$ thus $$8p^3 + 12p^2 + 6p + 1 = 2^n + 1 \implies 2p(4p^2 + 6p + 3) = 2^n$$ Note that $2p$ and $4p^2 + 6p + 3$ must be powers of $2$, however the latter is odd. Then, again $4p^2 + 6p + 3=1 \implies p=-1 \text{ or } p=-1/2$ both of which are impossible because $m > 0$