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.
As you learned from searching with Google, this is an elliptic curve. Integral points on elliptic curves are conveniently treated as a special subset of the rational points on such curves.
A lot of computation has already been done to identify the integral points and much else about elliptic curves with smallish coefficients. There turn out to be surprising equivalences among these, starting with changes of variables that are birational, so that $a,b$ can be rewritten as rational expressions in $x,y$ in a way that also allows $x,y$ to be expressed as rational expressions in $a,b$.
Here, setting $a = x/3$ and $b = y/9$ leads to the following Weierstrass form of your equation:
$$ a^3 = 3b^2 - 2 $$ $$ \frac{x^3}{27} = 3 \cdot \frac{y^2}{81} - 2 $$ $$ y^2 = x^3 + 54 $$
Note that if $(a,b)$ is an integral point of your original equation, then $x = 3a, y = 9b$ is an integer solution of the rewritten Weierstrass equation. In fact that equation is a special case of the form $y^2 = x^3 + k$ for integer $k$ known as a Mordell curve. It is known that there are only finitely many integral points on such curves.
In addition to instructions for software packages to analyze this curve, the results of what is known are summarized in this page, LMFDB Elliptic curve y² = x³ + 54. There we learn the only integral points are $(3,\pm 9)$.
It follows that the only integral points on your original curve are $(1,\pm 1)$.
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.
