Solve $x^3 +1 = 2y^3$ 
Solve $x^3 +1 = 2y^3$ in integers.

(Actually the original question was solve $x^n +1 = 2^{n-2} y^n$ but I can't even solve particular case $n=3$.)
Thanks in advance.

Mordell, Diophantine Equations, Chapter 23, Theorem 5 (page 203): If $d$ is an integer, $d\gt1$, there is at most one integer solution of $x^3+dy^3=1$ other than $x=1$, $y=0$. 
Also, Chapter 24, Theorem 5 (page 220): The equation $x^3+dy^3=1$ ($d\gt1$) has at most one integer solution with $xy\ne0$. This is given by the fundamental unit in the ring when it is a binomial unit, i.e., when the fundamental unit takes the form $x+y\root3\of d$. 
Both proofs are fairly long, and take some knowledge of Algebraic Number Theory. Maybe there's some elementary trick I'm not seeing for handling the case $d=2$. 
