No, it would float lower if anything. To see this think about the forces on the balloon:
- the acceleration due to gravity is $g$ and I assume this is constant (the planet is large, the balloon isn't getting very far up: this is a good assumption);
- the density of the gas inside the balloon is $\rho_H$, the density of the atmosphere is $\rho_A$.
If the 'radius' of the balloon is $r$ – this really means some characteristic linear size – then:
- the volume of the balloon is $k_1 r^3$;
- the surface area of the balloon goes like $r^2$ and I will assume that the mass of the balloon's structure can be assumed also to go like $r^2$ and we'll call it $k_2 r^2$.
$k_1$ and $k_2$ are just fudge factors to be determined, which depend on what the balloon is made of etc etc. Significantly $k_1 > 0$ and $k_2 > 0$.
So, OK now we can write an expression for the lift force generated by the balloon:
$$
\begin{aligned}
L &= g\left((\rho_A - \rho_H)k_1 r^3 - k_2 r^2\right)\\
&= g r^2\left((\rho_A - \rho_H)k_1 r - k_2\right)
\end{aligned}
$$
OK so what can we say about this? The first thing is that $g$ doesn't matter: it's just a factor outside the whole expression (in fact, $g$ does matter because it controls $P$, but it controls it the wrong way: other things being equal lower $g$ means lower $P$ and this is going to hurt us). The second thing is that we can assume $\rho \sim P$ where $P$ is atmospheric pressure to some reasonably good approximation (this comes from the ideal gas law: $PV - nRT$), so let's write $\rho_H \equiv \rho_{H,0}P$ & similarly for $\rho_A$:
$$L = gr^2 \left((\rho_{A,0} - \rho_{H,0})k_1rP - k_2\right)$$
And look at the expression $(\rho_{A,0} - \rho_{H,0})k_1rP - k_2$: in the first term there are a bunch of constants multiplied by $rP$ and in the second there is a constant.
This means that the lower $P$ is the smaller the lift is, and the more the fixed mass of the balloon starts to matter. In other words as $P$ goes down we get to fly less high. And, finally, as I said above, other things being equal, lower $g$ means lower $P$.
So balloons fly less high in lower gravity (or they need to be bigger so the factor of $r$ between the lift from the gas and the mass of the balloon's structure helps you more).
In fact, for hack value, we can go further than this: in the previous expression we had $P$, the atmospheric pressure, and $g$ the acceleration due to gravity. Well, if the mass of the atmosphere is $M_A$, the mass of the planet is $M_P$, and the radius of the planet is $R$, and the atmosphere is a thin layer (it is for rocky planets), then we get an expression for $g$:
$$g = \frac{G M_P}{R^2}$$
and also for $P$:
$$
\begin{aligned}
P &= g\frac{M_A}{4\pi R^2}\\
&= \frac{G M_P M_A}{4\pi R^4}
\end{aligned}
$$
And so the lift expression turns into
$$L = r^2\frac{G M_P}{R^2} \left((\rho_{A,0} - \rho_{H,0})
k_1 r \frac{G M_P M_A}{4\pi R^4} - k_2\right)$$
And I think that this expression is more useful than the previous one: it tells you that
- a bigger balloon helps (we know this);
- a more massive atmosphere helps;
- a more massive planet helps;
- a bigger planet hurts you quite badly;
- a larger value of $G$ helps.
This last thing I found extraordinary and I'm quite worried it means I've made a mistake.
