$2\mathbb{Z}$ is not ring isomorphic to $3\mathbb{Z}$, But Why? (Verify Me)

$x^2 - 2x = x(x-2)$ is not a valid factorization in $3\mathbb{Z}$ precisely because $2 \notin 3\mathbb{Z}$.

But you can just directly conclude that $x^2 - 2x = 0$ has no solutions in $3\mathbb{Z}$.

Namely, if $x = 3k$ then

$$9k^2 = x^2 = 2x = 6k \implies 9k= 6$$

which is a contradiction.

Seeing everything as sub(non unital)rings of $\mathbb{Z}$ makes your argument work.

But you can also write your equation as $x^2=x+x$, and plug in $3k$, and see that it has no solution.

Your argument is correct until $x^2 - 2x = 0$ which suffices to complete the proof. Here is another way to see this:

Your ring isomorphism $\phi$ is an isomorphism of abelian groups which are both infinite cyclic. Hence the generator $2$ of $2\mathbb{Z}$ is mapped to a generator of $3\mathbb{Z}$, i.e. $\phi(2) = \pm3$. This implies $\pm 6 = \phi(2) + \phi(2) = \phi(2 + 2) = \phi(2 \cdot 2) = \phi(2) \cdot \phi(2) = (\pm 3) \cdot (\pm3) = 9$, a contradiction.