<s>[INST] Integer solutions for $x^3+2=y^2$? I've heard a famous result that $26$ is the only integer, such that $26-1=25$ is a square number and $26+1=27$ is a cubic number.In other words, $(x,y)=(5,3)$ is the only solution for $x^2+2=y^3$.
How if we make it like this $x^3+2=y^2$? Are there any integral solutions? If so, finite or infinite many?
I've checked first $100$ naturals and no solutions satisfy the equation. However, I have no ideas how to start the proof.
 [/INST] Edit: Misread the question. Will leave the answer for a while, since the writeup of the solution to the wrong problem  may be interesting to MSE users. 
We need to take a moderately lengthy detour through the arithmetic of $\mathbb{Z}[\sqrt{-2}]$, that is, the arithmetic of numbers of the form $a+b\sqrt{-2}$, where $a$ and $b$ are integers.
It turns out that the arithmetic is "nice," a version of the Unique Factorization Theorem holds. That crucial part of the work is not done in this answer.  
The rest of the proof is quite natural. We do that "rest," in detail. 
Suppose that our equation holds. Then $(x+\sqrt{-2})(x-\sqrt{-2})=y^3$. Any non-trivial common divisor of the two terms on the right must divide $2\sqrt{-2}$. But it is clear that $x$ is odd, for if $x$ is even then $x^2+2\equiv 2\pmod{4}$, so $x^2+2$ cannot be a perfect cube. Thus $x+\sqrt{-2}$ 
and its conjugate each have odd norm. But any non-trivial divisor of $2\sqrt{-2}$ has even norm. So $x+\sqrt{-2}$ and $x-\sqrt{-2}$ are relatively prime in $\mathbb{Z}[\sqrt{-2}]$.
By Unique Factorization, each of $x+\sqrt{-2}$ and its conjugate have the shape $\epsilon w^3$, where $\epsilon$ is a unit. The only units are $\pm 1$. If $\epsilon=-1$, it can be absorbed into $w$, so we may assume that 
$x+\sqrt{-2}=w^3$. 
Let $w=a+b\sqrt{-2}$. Expanding the cube, we find that
$$x+\sqrt{-2}=(a+b\sqrt{-2})^3= a^3-6ab^2+(3a^2b-2b^3)\sqrt{-2}.$$
This yields the equations $a^3-6ab^2=x$ and $3a^2b-2b^3=1$.
But $3a^2b-2b^3=1$ is a very restrictive condition. Since $b$ divides the left side, we must have $b=\pm 1$. If $b=1$ we get $a=\pm 1$, while if $b=-1$, there is no $a$ that works.
Thus $x=a^3-6ab^2$ with $b=1$ and $a=\pm 1$. That gives the solutions $x=\pm 5$, $y=3$. 
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