 Solution

Use countability of negative curves. A smooth projective surface over $\mathbb{C}$ is a compact smooth manifold homotopy equivalent to a finite CW complex, so its integral cohomology group $H^2(X,\mathbb{Z})$ is finitely generated and therefore countable. Distinct irreducible curves $C,D$ with $C^2<0$ cannot have the same cohomology class: that would give $C\cdot D=C^2<0$, while distinct effective curves have nonnegative intersection multiplicities. Assigning a negative curve its cohomology class is thus injective, proving there are at most countably many negative curves.

For zero self-intersection, uncountable families do occur. On the product of two projective lines ${\mathbb{P}}^1\times {\mathbb{P}}^1$, every fiber ${\mathbb{P}}^1\times \{p\}$ has self-intersection zero, since a different fiber is disjoint and has the same numerical equivalence of divisors class. The parameter $p\in {\mathbb{P}}^1(\mathbb{C})$ varies over an uncountable set. Thus yes, there can be uncountably many such curves.