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.
I will leave it to others to find an elementary proof, but I just wanted to say that this equation defines an elliptic curve $E:y^2=x^3+2$. The Mordell-Weil theorem states that the set of rational points is a finitely generated abelian group, so $$E(\mathbb{Q})\cong T_E \times \mathbb{Z}^{R_E},$$ where $T_E$ is a finite subgroup, and $R_E\geq 0$ is called the rank of the elliptic curve. There are methods to calculate $T_E$ and $R_E$ (such as the Nagell-Lutz theorem, or the method of $2$-descent), and these methods are implemented in software such as Sage and Magma. In this case: $$T_E=\{\mathcal{O}\}$$ where $\mathcal{O}$ is the point at infinity with projective coordinates $[1,0,0]$, and $R_E=1$, with a generator $(-1,1)$. So the set of all rational points on $E$ is: $$E(\mathbb{Q})=\{nP: P=(-1,1)\}.$$ The multiples of $P$ have the following coordinates: $$2P=(17/4,-71/8),\ 3P=(127/441, 13175/9261),$$ $$4P=(66113/80656, -36583777/22906304),\ldots$$ Using the theory of heights one can show that $P=(-1,1)$ and $-P=(-1,-1)$ are the only integral points on $E$, but I will not show this here.
I calculated the torsion and rank using the online Magma calculator:
E:=EllipticCurve([0,0,0,0,2]);
Rank(E);
TorsionSubgroup(E);
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 $x=\pm 5$, $y=3$.
As mentioned in the comment above, this is a particular elliptic curve called the Mordell curve. If we assume that $k$ is an integer then the Birch-Swinnerton-Dyer conjecture says that it has finitely many solutions if and only if the $L$-function of the elliptic curve does not vanish at $s=1$
