Can we drop smoothness in uniformity conjecture if we don't count singular points?

The uniformity conjecture basically states that the number of rational points on a smooth curve of genus $g >1$ over number field is bounded.

If we drop smoothness, there is counterexample coming from unbounded number of singular points.

Can we drop smoothness if we don't count singular points?

i.e., The number of rational points which are not singular is bounded in term of the genus $g > 1$ and the number field?

• Arithmetic genus or geometric genus? Commented Oct 15, 2017 at 11:39
• @JoeSilverman geometric as is commonly accepted to write.
– joro
Commented Oct 15, 2017 at 11:40

Yes. Consider the smooth projective model $\tilde{C}$ of your (geometrically irreducible) curve $C$ of (geometric) genus $g \ge 2$ over a number field $k$. Then the nonsingular points on $C$ are in bijection with a Zariski-open subset of $\tilde{C}$, and this bijection preserves $k$-rational points (because the canonical morphism $\tilde{C} \to C$ is defined over $k$). So $C$ can have at most as many non-singular $k$-rational points as $\tilde{C}$ has $k$-rational points, and the latter number is conjecturally bounded in terms of $g$ (which is the genus of $\tilde{C}$) and the field $k$ (or just its degree over $\mathbb Q$).