Does this Diophantine cubic have solutions? I'm interested in whether
$$
2x^3-1=z^3
$$
has any positive integer solutions. It has solutions mod $m$ for $m\le10^7$ so it seems unlikely that modular means would suffice to prove that there are not solutions. On the other hand I can't find any solutions for $x<10^9$ either.
I'm sure this is a routine problem but I don't know the techniques. Help?
Application: I was reading about near-Fermat triples $x^3+y^3=z^3+1$ where the restriction $x<y<z$ was given rather than $x\le y<z$ and I wondered if the excluded case was possible.
 A: Euler gave an elementary proof that there are no non-trivial solutions. (Actually, he solved the more general equation $y^3+z^3=2x^3$.)
Having asked this before (see Rational solutions of $x^3+y^3=2$) I was pointed to Chapter II, pg 78 here: http://matwbn-old.icm.edu.pl/kstresc.php?wyd=10&tom=42&jez=en.
If you are interested to (partially) solve the equation yourself, the substitutions made in the beginning might allow you to proceed the proof.
A: Some points:
Greatest common divisor: If $g=gcd(x,y)$, from $1=2x^3-z^3$ we know that $g^3|1$, or $g=1$ and $x$ and $y$ are relative prime.

Parity: Since $2x^3$ is even and 1 is odd, $z^3$ is odd and $z$ is odd.

Factors of z: If  $p|z$, so $p^3|z^3=2x^3-1$ hence $2x^3\equiv 1 \pmod {p^3}$. Computer search shows $y$ is not dividable by any of $\{2, 3, 7, 13, 19, 37, 61, 67, 73, 79, 97\}$.

Solving mod primes using computer: From this I found some results that may help in solving or brute-forcing the problem: $x\equiv 0,1,2 ,4 \pmod7$, $x\equiv 0,1,3,9 \pmod{13}$, $x\equiv 0,1,7,11\pmod{19}$, 
