Two variables Diophantine equation $x^3 - 2y^2 =54$ How  can  we  solve  $x^3 - 2y^2 =54$  Diophantine  equation  in  natural  numbers? Modular  arithmetic  method  is  not  useful  maybe?

Since $x$ must be even, $x=2z$, we have:
$$ 4z^3- y^2 = 27 \tag{1}$$
or:
$$z^3 = \frac{y+3\sqrt{-3}}{2}\cdot\frac{y-3\sqrt{-3}}{2}=w\cdot\overline{w},\tag{2}$$
where $w$ lies in the ring of integers of $\mathbb{Z}[\sqrt{-3}]$, that is a UFD.
 This implies that, if $w$ and $\overline{w}$ are coprime, $w=\frac{y-1}{2}+3\omega$ must be a cube in the Eisenstein integers $\mathbb{Z}[\omega]$ (notice that $y$ must be odd, so $\frac{y-1}{2}\in\mathbb{Z}$), but:
$$ (a+b\omega)^3 = a^3+3a^2 b\omega + 3ab^2\omega^2 +b^3 = (a^3+b^3-3ab^2)+3ab(a-b)\omega\tag{3}$$
and $3ab(a-b)=3$ has no integer solutions. The only possibility left is that $w=\sqrt{-3} u^3$ and $\overline{w}= -3v^3$ with $\gcd(u,v)=1$. This happens for $(x,y)=(6,9)$.
