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.

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.
