Positive integer solution to $a^3=3b^2-2$ The equation $a^3=3b^2-2$ seem to only have one positive integer solution $(a,b)=(1,1)$, but I am unable to prove that.
What I did:
Google told me that the elliptic curve theory might help me, but I don't really know anything about it.
This thread in MO looks similar to what we have, but I was unable to apply it in my case.

Let $a$ and $b$ be integers such that $a^3=3b^2-2$. First note that $a$ is not divisible by $3$, and that $a$ and $b$ are both odd. In the UFD $\Bbb{Z}[\sqrt{6}]$ we have the factorization
$$3a^3=9b^2-6=(3b+\sqrt{6})(3b-\sqrt{6}),$$
where the gcd of the two factors divides their difference $2\sqrt{6}$ as well as their product $3a^3$. As noted $a$ is odd, and so the two factors are coprime. The prime factorization $3=(3+\sqrt{6})(3-\sqrt{6})$ in $\Bbb{Z}[\sqrt{6}]$ shows that, after changing the sign of $b$ if necessary, we have
$$3b+\sqrt{6}=u(3+\sqrt{6})(c+d\sqrt{6})^3,\tag{1}$$
for some unit $u\in\Bbb{Z}[\sqrt{6}]$ and some integers $c$ and $d$. The unit group of $\Bbb{Z}[\sqrt{6}]$ is generated by $-1$ and $5+2\sqrt{6}$, meaning that $u=\pm(5+2\sqrt{6})^k$ for some integer $k$. Adjusting $c$ and $d$ if necessary, we may assume without loss of generality that $k\in\{-1,0,1\}$ and that we have the $+$-sign. Then we are left with solving
$$3b+\sqrt{6}=(5+2\sqrt{6})^k(3+\sqrt{6})(c+d\sqrt{6})^3,$$
for some integers $c$ and $d$, and some $k\in\{-1,0,1\}$. Expanding the product for each of the three values of $k$ and comparing coefficients of $\sqrt{6}$, we end up with the following three Diophantine equations:
\begin{eqnarray*}
1&=&-c^3+9c^2d-18cd^2+18d^3\\
1&=&c^3+9c^2d+18cd^2+18d^3\\
1&=&11c^3+81c^2d+198cd^2+162d^3\\
\end{eqnarray*}
These are three cubic Thue equations, for which there exist effective methods for finding all integral solutions. The first has the unique solution $(c,d)=(-1,0)$, the second has the unique solution $(c,d)=(-1,0)$, and the third has no integral solutions. These two solutions correspond to $b=-1$ and $b=1$, respectively, and this shows that these are the only integral solutions.
