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.
 A: 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 $d=2$. 
A: I've only recently joined Maths SE, so didn't see the question when originally posted.
If $x^3 + 1 = 2y^3$ then $1 (= 1^3), y^3, x^3$ are in arithmetic progression.  However, Y. Hellegouarch, in Introduction to the Mathematics of Fermat-Wiles (English translation, Academic Press 2002) states the following on p 342:
Denes Conjecture: Let p be an odd prime.  If the three natural non-zero integers $x^p, y^p, z^p$ lie in an arithmetic progression, then x = y = z.
It is stated on p 343 that this conjecture has been proved by Darmon & Merel.  The reference is to the following (which I have not seen): Darmon H. and Merel L., Winding quotients and some variants of Fermat's last theorem  J. Reine Angew. Math 490 81-100, 1997.
This implies that the above equation has no non-trivial solution in positive integers, and more generally that $x^3 + z^3 = 2y^3$ has no non-trivial solution in positive integers.  It would seem however to leave open the possibility of a solution with negative x and y.
Addendum:
I realise that the Darmon & Merel result is not necessary to show that $x^3 + z^3 = 2y^3$ has no non-trivial solution in integers. The impossibility is proved in Chapter 2, on p 79 of Sierpinski Elementary Theory of Numbers which I found could be accessed here.  This covers the case of negative x and y too.
A: For a classical proof by Euler: http://books.google.es/books?id=mqI-AAAAYAAJ&hl=es&pg=PA234#v=onepage&q&f=false
A: Here's a sketch of what I would do. 
Instead of considering the equation $x^3 + 1 = 2y^3,$ I would look at the equation 
$$(-x)^3 + 2y^3 = 1.$$
Now what this says is that the pairs $(x,y)$ which solve your equation are in one to one correspondence with elements of the form $-x + \sqrt[3]{2}y$ in $\mathbb{Q}(\sqrt[3]{2})$ of norm $1$ over $\mathbb{Q}$ where $x,y\in \mathbb{Z}.$ So all solutions can be obtained by examining the unit group of the ring of integers of $\mathbb{Q}(\sqrt[3]{2}).$
Note by Dirichlet's unit theorem $\mathcal{O}_{\mathbb{Q}(\sqrt[3]{2})}^{\times} \cong \mathbb{Z}/2 \oplus \mathbb{Z}.$ Furthermore, the free part of this unit group is isomorphic to elements of norm $1.$  Now the problem is 'easy.' Find a primitive unit for this unit group and examine which powers yield elements of the desired form. Good luck! 
