The material I'm studying has the solution for $m^2 = n^3 + 4$ but I can't quite understand it. It starts like this:

First suppose, for a contradiction, that $m$ is odd. Then $n^3 = m^2 − 4 = (m + 2)(m − 2)$. Any common factors of $m + 2$ and $m − 2$ divide $(m + 2) − (m − 2) = 4$ and so, $\gcd(m + 2, m − 2) = $1$.

It follows that $m + 2$ and $m − 2$ are both cubes.

a) How did they get to the conclusion $\gcd(m+2, m-2)=1$? I mean, at this stage of the proof, I don't know.

$m\neq 10$ so I could think if $m=10$ then gcd(12,8) $=4\neq 1$.

b) Why does it follow that $m+2$ and $m-2$ are both cubes, and not that their multiplication is a cube?

Let's carry on the proof.
[...] No two odd cubes differ by 4, giving a contradiction.
Since $m$ is even there exists $m'$ such that $m = 2m'$. We thus have that $4m'^2 = n^3 + 4$ and $n$ is divisible by $2$ so we write $n = 2n'$. Now, $m'^2 = 2n'^3 + 1$ and so $m'$ is odd so let $m' = 2r + 1$. Substituting this into the previous equation tells us that $2r^2 + 2r = n'^3$.

No problems with the above, let's carry on

Finally,$r^2 + r = r (r + 1)$ and since $r$ and $r + 1$ are coprime we must have that one of $r$ and $r + 1$ is a cube and the other is four times a cube.

c) Why are we considering $r^2+r$ when we have $2r^2+2r=n'^3$? Yes, they have a common factor of $2$, but we don't know this about $n'$ do we? we can't cancel them.

d) ``since $r$ and $r + 1$ are coprime we must have that one of $r$ and $r + 1$ is a cube and the other is four times a cube'' ????

Any help with the above is really appreciated.