Prove that $2^{n}+1$ is not a cube of an integer for all $n\in\mathbb{N}$

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 $6\pmod 7$.

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. :-)

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.