Finding integer solutions to $y^2=x^3-2$

I have the equation:
$$y^2=x^3-2$$
It seems to be deceivingly simple, yet I simply cannot crack it. It is obviously equivalent to finding a perfect cube that is two more than a perfect square, and a brute force check shows no solutions other than $y=5$ and $x=3$ under 10,000.
However, I can’t prove it.

Are there other integer solutions to this equation? If so, how many? If not, can you prove that there aren’t?

Bonus: What about the more general equation”
$$y^2=x^3-c$$
Where $c$ is a positive integer?

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The only integral solutions to your first problem are $(3, \pm 5)$. The general class of equations are known as Mordell’s equation. A fairly elaborate discussion and case by case analysis is provided here.

Fact : $\mathbb{Z}[\sqrt{-2}]$ is Unique factorization domain.
lemma : in every UFD, if the product of two numbers, which are relatively prime is a cube, then each of them must be a cube.
There is no any solution

$$x^3 = (y+\sqrt{-2})\times(y-\sqrt{-2})$$

the greatest common divisor of these factors will divide 2-times the $\sqrt{-2}$,
which is lead to only finitly many cases. (some cases can be shown impossible, only by the modular an congrunce arithmetic.)

finally we have:
$$y+\sqrt{-2}=(a+b\sqrt{-2})^3$$
which lead us to the system of equations as follows:
$$a^3-6ab^2=y$$ and
$$b^3+3a^2b=1$$
then
$$b(b^2+3a^2)=1$$
which implies the assertion.