Suppose $a^3 + b^3 = c^3, \ a,b,c \in \mathbb Z^+\ \ (1),\ $ we can assume: $\ a,b$ and $c$ are coprime.Decompose $c^3 - b ^ 3$ into $(c - b)((c - b) ^ 2 + 3cb) = a ^ 3\ $give us three cases $\ C_{1},C_{2} \ \text{and} \ C_{3}$
$\quad C_{1}\ c - b = a ^ 3,\ $ not reasonable.
$\quad C_{2}\ 1 < c - b < a ^ 3$
$\quad C_{3}\ c - b = 1$
$\quad$In Case $C_{2},\ $ we see $(c - b) \mid a ^ 3,\ $then there exists a number $m>1$ divides $(c-b)$ and $a,\ $and we see $m \mid 3cb$
$\quad \quad C_{21}\ m=3,\ m \not\mid c,b$
$\quad \quad C_{22} \implies (c \lor b,a)>1,\ $conflict our assumption.
$\quad \quad $In Case $C_{21}, \ c-b\,$ will of the form $3^k\,$ on the two sides requires $\,k = 3 u - 1\,$ and $\,a = 3^{u} \, v\,$(contributed by MSE user dxiv). We decompose $c^3 - a ^ 3$ also give us three cases
$\quad \quad \quad C_{211}\ c - a = b ^ 3,\ $ not reasonable
$\quad\quad \quad C_{212}\ 1 < c - a < b ^ 3,\ $ then there exists another $m$ conflict our assumption because of $3$ has already divides $a$
$\quad\quad \quad C_{213}\ c - a=1,\ $ then we have $(c-1)^3 + (c-3^k)^3=c^3,\ c\ $ must be even. Decompose $a^3 + b ^ 3$ into $(a + b)((a + b) ^ 2 - 3ab) = c ^ 3\ $give us two cases
$\quad \quad \quad \quad C_{2131},\ a + b = a ^ 3,\ $ not reasonable.
$\quad \quad \quad \quad C_{2132},\ $ if there does not exists another $m$ conflict our assumption then while $c=2^nq$ for some odd $q,\ a+b$ must be equals to $2^{3n},\ $ then we have $a+b=(c-1)+(c-3^k)=2^{3n},\ $the question will be: let t = $ \dfrac{2^{3n}+3^k+1}{2^{n+1}},\ n ,\ k \in \mathbb Z^+\ \ \ (2),\ $ if $t$ is an integer, prove $3$ divides $t\ $(proof see aditional).
$\quad$In Case $C_{3}$, we will have the case same to $C_{213},\ $the only difference is $a,b$ swap position.
Here all cases return false, so $(1)$ is not tenable.
My question is I can't believe the problem is that simple, are there something wrong in my reasoning?
Aditional: MSE user (another Mike)Mike's proof of question $(2)$
HINT: For each integer $k$, note that $(3^k+1)\pmod 8$ is either $2$ [if $k$ is even] or $4$ [if $k$ is odd].
If $(3^k+1) \pmod 8$ is $2$, then for $t$ to be an integer then, note that $n$ cannot be a positive integer. So $t$ is never a positive integer if $(3^k+1)\pmod 8$ is $2$.
If $(3^k+1)\pmod 8$ is $4$ then for $t$ to be an integer, $n$ must be $1$. And so $3n$ must be $3$. But then what do you know about $2^3+1$? What divides $2^3+1$?