A tricky equation

Hey guys! First blog post. I wanna share with you, the story of a seemingly easy but startlingly complex equation I stumbled upon.

$$y^2=x^3-2$$

To add to it, this was in our rising 8th 'math placement' worksheet, right next to a question about finding the area of a square with a width of 5. So, naturally, I thought it was going to be easy. But upon further inspection, after attacking it with many tools of Algebra 1, I hadn't even made a dent.

It is hard not to show that the question is equivalent to finding a perfect square that is 2 less than a perfect cube (I should mention we are looking for non negative integral solutions). Well, immediately, a solution sprang to mind. $(3,5)$. $3^3=27$, $5^2=25$. Looking back, I think that's all the teacher wanted us to do. Find that solution by trial and error. But, was it the only solution?

Looking back, I could have spared myself a lot of time and energy if I just assumed that solution was the only one and moved on with my life. But that's not how we roll :)
So, the first thing to do was an exhaustive search. I checked it up to $100,000$, and it seemed like there were no other solutions. But remember, a pattern is not a proof. Remember Richard Guy's conjecture that $\gcd(n^{17}+9, (n+1)^{17}+9)$ is always 1? Well, the first counterexample is $$n=8424432925592889329288197322308900672459420460792433$$.
So yeah, an exhaustive search is not a proof. We are here for mathematical rigor.

Well, first, let's establish some preliminary results.
Assuming there is an $x$ and $y$ other than $(3,5)$, both x and y would have to be odd.
Proof:
Assume $x$ is even. Then, $x$ can be expressed as $2z$ for some integer z. So $x^3$ can be expressed as $8z^3$ Rearranging the original equation with these new terms, we get:
$$y^2+2=8z^3$$
And it is, from that, obvious that $y^2+2$ is a multiple of 8. We can also express this as a modular congruence. But before I do that, let me explain what modular congruence is.

For a positive integer $n$, two integers $a$ and $b$ are called congruent if their difference $a-b$  is a multiple of $n$. This is notated:
$$a \equiv b \pmod{n}$$
Notice that is not an equals sign, it's 3 bars.
So, we know the difference between $y^2$ and $-2$ is a multiple of 8. We can notate this:
$$y^2 \equiv -2 \pmod{8}$$
However, $-2 \pmod{8}$ is not a perfect square, so $y^2$ is not even, which implies $x^3$ is not even (Because $x^3$ is $y^2+2$ and and odd number plus two is always odd), which implies both $x$ and $y$ are not even.
So, we have established that both $x$ and $y$ are odd. Now what? Well, here is where it gets a bit tricky.

First, from now on we will be working with the equivalent equation:
$$y^2+2=x^3$$
Which was gotten by simply adding 2 to both sides.
We can factor this equation quite easily:
$$x^3=(y+\sqrt{-2})(y-\sqrt{-2})$$

Now, at this point, I got stuck. I had no idea what do from here. I tried a bunch of different stuff, but nothing worked. But after a few days of thinking, some help from the Math Stackexchange + some personal stupidity worked out.

I am going to prove that $(y+\sqrt{-2})$ and $(y-\sqrt{-2})$ are relatively prime (i.e they share no factors) within $Z[\sqrt{-2}]$. Now, to understand what that means, consider the set of all integers. Well, we can do operations outside of just the numbers. $\mathbb{Z}[\sqrt{-2}]$ is  the set of all numbers of the form $(m+n\sqrt{-2})$ where $m$ and $n$ are integers. So, to say that $(y+\sqrt{-2})$ and $(y-\sqrt{-2})$ are relatively prime in $\mathbb{Z}[\sqrt{2}]$ means they have no common factors within that set other than 1)
To prove this, let us consider a variable $d$ which is a common divisor. $d$ divides their difference: $$(y+\sqrt{-2})-(y-\sqrt{-2})=2\sqrt{2}$$
Which means $N(d)$ divides $N(2\sqrt{2})=8$* but $N(d)$ also divides $N(y+\sqrt{-2})=y^2+2$, which is odd because $y$ is odd, as we established.  And the only factor of 8 that also divides odd numbers is $1$. So because $N(d)$ is 1, $d$ is a unit in $\mathbb{Z}[\sqrt{-2}]$. So, $(y+\sqrt{-2})$ and $(y-\sqrt{-2})$ are both unit multiples of $+1$ and $-1$. Those are both cubes, so $(y+\sqrt{-2})$ and $(y-\sqrt{-2})$ are both cubes. We can express it this way:

$$y+\sqrt{-2}=(m+n\sqrt{-2})^3$$
For some integers $m$ and $n$.
Distributing and subtracting the square root gives us:
$$y = m^3 − 6mn^2 = m(m^2 − 6n^2 ), 1 = 3m^2n − 2n 3 = n(3m^2 − 2n^2 )$$
And from this we can extract solutions, with the help of Mathematica.
From the second equation, $n=\pm{1}$. When $n = 1$ the second equation says $1 = 3m^2 − 2$, so $m = \pm1$. Then $y = \pm1(1 − 6) = ±5$ and $x^3 = y^2 + 2 = 27$, so we recover the solutions $(x, y) = (3, \pm5)$. When $n = −1$ we have $1 = −(3m2 − 2 \cdot 1^2 ) = −(3m2 − 2)$, so $1 = 3m^2$ , which has no solution in $\mathbb{Z}$. (Thanks to Andre Nicolas on MSE for pointing this out)

So boom, there we go! We have extracted the solutions from this equation. I'm not sure it's what our math teacher wanted us to do... but, it was fun in any case :) See you in the next post!

*$N(m+n\sqrt{D})=a^2+Db^2$
Note: To actual people who know a lot more math than me, this might seem like a very shaky proof. I'm putting out a lot of assumptions and not really explaining a lot of things in depth. This is because I didn't want to get too into unique factorization domains and modular arithmetic. If you do want to discuss it in depth, feel free to email me!
Also, a lot of it is probably wrong! I had mathematica check most of it for me, but diving straight into something like this with no experience probably means my logic is flawed in places. For