What does this converge to and why?

When we ask "What does this expression converge to?" we usually mean "If you cut off this infinite expression after a certain point, you get a number. As you cut off the expression later and later, that number gets closer and closer to some limiting value. What is that limiting value?"

However, with this continued fraction, we could truncate it in two ways: after a $2$, getting something like $$ \cfrac{2}{3 - \cfrac2{3 - \cfrac2{3- \cfrac2{3 - 2}}}}, $$ or after a $3$, getting something like $$ \cfrac{2}{3 - \cfrac2{3 - \cfrac2{3- \cfrac2{3 - \cfrac23}}}}. $$ With many continued fractions, both of these result in the same answer. Here, however, cutting the expression off when the number "at the end" is a $2$ always means that the whole thing simplifies to $2$; cutting it off when the number at the end is a $3$ gives a value approximately equal to $1$ (more precisely, it gives $\frac{2^{n+1}-2}{2^{n+1}-1}$, where $n$ is the number of $3$'s used, and this converges to $1$ as $n \to \infty$).

You could decide that the first kind of truncation is the kind you want, in which case the continued fraction converges to $2$. Or you can decide that the second kind of truncation is the kind you want, in which case the continued fraction converges to $1$. (Usually, the second kind of truncation gets used by convention for continued fractions.)

You can even decide that both kinds of truncations are fine, in which case the continued fraction does not converge: it alternates between values equal to $2$ and values close to $1$. This is the option that we usually pick for the sum $1 - 1 + 1 - 1 + 1 - 1 + \dots$, for instance: this has values of $1$ or $0$ depending on if we cut it off at an even or odd position, but there's no reason to pick one over the other, so we say it does not converge.

I think you need to question the assumption that you need to reject a solution.

You're probably thinking:

But this problem clearly only has one solution!

But why? Didn't you just prove $x = \dfrac{2}{3-x}$ has two solutions?

But that's an implicit equation. This is an explicit expression. An expression can't equal 2 things!

Sure it can. $\sqrt{-1}$ is an expression, and only equal to $i$ because we arbitrarily decided to ignore $-i$.
But if you want to ignore things arbitrarily, you can do that here too—that's just a preference.

But, like, come on. If you chop off this expression, it'll clearly only have one value. So if you keep chopping it off later and later, clearly it can't converge to two values?!

Oh, so you want convergence? In that case, there's no reason to believe there's exactly 1 solution. In fact, you're lucky if you get any solution. But if that's what you want, see the other answers here.

The fundamental problem here is that you have not defined the "$=$" operator between a finite value and expressions that have an infinite number of terms.

In fact, there are multiple definitions, and depending on which you choose, it may make sense to reject none, some, or all of the solutions.

You can look at the stability of the solutions. You start with some $x_0$ close to one of your solutions, and check if $$x_1=\frac{2}{2-x_0}$$ is closer to that solution. So if $x_0=1+\epsilon$, where $|\epsilon|\ll 1$, we have $$x_1=\frac{2}{3-(1+\epsilon)}=\frac{2}{2-\epsilon}\approx\left(1+\frac{\epsilon}{2}\right)$$ This is closer to $1$ than $x_0$, so the solution is stable.

We repeat the procedure for $x_0=2+\epsilon$. $$x_1=\frac{2}{3-(2+\epsilon)}\approx2(1+\epsilon)=2+2\epsilon$$ This $x_1$ is further from $2$ than $x_0$, so the solution is not stable.