Prove that $2^{n}+1$ is not a cube of an integer for all $n\in\mathbb{N}$ Prove that $2^n+1$ is not a cube for any $n\in\mathbb{N}$.
I managed to prove this statement but I would like to know if there any other approaches different from mine.
If existed $k\in\mathbb{N}$ such that $2^n+1=k^3$ then $k=2l+1$ for some $l\in\mathbb{N}$. Then $(2l+1)^3=2^n+1 \iff 4l^3+6l^2+3l=2^{n-1}$. As I am looking for an integer solution, from the Rational Root Theorem $l$ would need to be of the form $2^j$ for $j=1,...,n-1$. But then
$$4(2^j)^3+6(2^j)^2+3\times2^j=2^{n-1} \iff 2^{2j+2}+3(2^{j+1}+1)=2^{n-1-j}$$
the LHS is odd which implies that $j=n-1$. Absurd.
Thank you in advance.
 A: Here is a parity-based solution that avoids the rational root test.
If $2^n+1=m^3$, then $2^n=m^3-1=(m-1)(m^2+m+1)$, so $m-1=2^k$ for some $k\le n$, and
$$2^n+1=\left(2^k+1\right)^3=2^{3k}+3\cdot2^{2k}+3\cdot2^k+1\,.$$
Then $2^n=2^k\left(2^{2k}+3\cdot2^k+3\right)$, so $2^{n-k}=2^{2k}+3\cdot2^k+3$ is odd and greater than $1$, which is impossible.
Added: As one can see from the comments below, there are many ways to continue this argument after the first line. I took what I think of as the follow-your-nose approach, i.e., the most obvious, straightforward one, not necessarily the neatest. (And speaking of neatest, I quite like the one by rtybase.) Then again, folks’ noses don’t always point in the same direction. :-)
A: Invoking an argument more powerful than needed for this:
there cannot be any solutions to $2^n+1=m^3$ (i.e., $m^3-2^n=1$) by Mihăilescu's theorem,
which states that  $2^3$ and $3^2$ are the only two powers of natural numbers
whose values are consecutive.
A: Here is a different approach.
Modulo $7$, there aren't so many cubes, so that can be a good setting to investigate such problems:
$2^n+1\equiv 2, 3, $ or $5\pmod7$, but $m^3\equiv0, 1, $ or $6\pmod 7$.
A: Suppose $2^n + 1 = k^3$. Then $2^n = k^3 - 1 = (k^2 + k + 1)(k - 1)$. So both factors are even ($k = 2$ doesn't work; the first factor is at least $3^2 + 3 + 1 = 13$, it can't be 1). But the first factor is always odd, contradiction.
A: Let $$2^n=m^3-1\\\implies 2^n=(m-1)(m^2+m+1)\\\implies(m-1)=2^a\text{ and }(m^2+m+1)=2^b\\\implies3m=(m^2+m+1)-(m-1)^2=2^b-2^{2a}$$ Now, since $m$ is odd, we must have $a=0$ or $b=0$. But $(m-1)<(m^2+m+1)$ implies $a=0$. This implies $m=2$ a contradiction since $m$ must be odd.
A: Let's set the cubes to $8m^3$ and $8m^3+12m^2+6m+1$. As $8m^3$ is even and it doesn't works for $n=0$, that's impossible. For the second one, ignoring the $1$ you can factor it to $2m(4m^2+6m+3)$. Since there ain't any natural in which $4m^2+6m+3=1$ it is impossible to be a $2^n$ for natural $n$.
