Fick's law as written—i.e., that matter tends to move to eliminate concentration (or density) differences—ignores gravity. As you note, this law would predict that any density difference should spontaneously undo itself, which of course doesn't happen; the equilibrium density is higher at the bottom of a fluid.
More fundamentally, matter tends to move to eliminate chemical potential differences. (The chemical potential is like a sophisticated version of concentration that incorporates chemical interactions and can be modified to incorporate body forces such as gravity.) Let's expand the scope to this more general framework to shed any Fick's law assumptions that are inconsistent with the current scenario.
The relevant potential I'm going to work with is modified from the Gibbs free energy $G\equiv U-TS+PV$, which is appropriate for systems in thermal and mechanical contact with their surroundings, as with the fluid you mention. Assuming a constant temperature $T$ and amount of fluid $N$, the Gibbs free energy in differential form simplifies to
$$dG=V\,dP,$$
meaning simply that I can increase the energy of a region by increasing the pressure $P$. To this I'll add an additional term $\boldsymbol{mg\,dz}$, as I could also increase the energy of a region with mass $\boldsymbol{m}$ by increasing its height $\boldsymbol{z}$ in our now-relevant gravity field $\boldsymbol{g}$:
$$dG^\star=V\,dP+mg\,dz,$$
where $G^\star$ is the modified Gibbs free energy that incorporates gravity. In hydrostatic equilibrium, we expect this potential to balance everywhere: $dG^\star=0$. We obtain
$$dP=-\rho g\, dz,$$
with density $\rho\equiv\frac{m}{V}$. For nearly incompressible materials such as fluids, $\rho$ is nearly constant, so integration recovers the hydrostatic formula $\Delta P=-\rho g\Delta z$, as expected. For ideal gases, where $P=\frac{\rho}{M}RT$ (molecular weight $M$, gas constant $R$), we obtain the isothermal barometric formula $P=P_0\exp(-Mgz/RT)$, as expected. This indicates that the additional gravitational potential energy term is doing its job.
So what do Fick's laws of diffusion look like in the presence of gravity? This question is relevant when considering the solid-state diffusion of heavy impurities, for example, which tend to sediment toward the bottom. We note that the chemical potential $\mu$ is just the partial molar Gibbs free energy:
$$\mu\equiv\left(\frac{\partial G}{\partial N}\right)_{T,P}.$$
Gradients in potential fields tend to drive directional flow $\boldsymbol{J}$ that eliminates the gradient. Fick's first law is a linear approximation to this behavior:
$$\boldsymbol{J}=-L\nabla \mu,$$
where $L$ is some linear-response coupling coefficient.
The chemical potential is related to the concentration $C$ by
$$\mu=\mu_0+RT\ln \gamma C,$$
where $\mu_0$ is a reference zero and $\gamma$ is the activity coefficient, which captures information about chemical interactions. In ideal solutions, as assumed in Fick's laws, $\gamma=1$. (In the more general case, unmixing phenomena such as oil–water separation can be studied, even though Fick's laws always predict complete mixing.)
So we have
$$\boldsymbol{J}=-\frac{LRT}{C}\nabla C,$$
but if we add the constraint that impurity movement in a lattice generally requires equal host movement in the opposite direction, the concentration C in the denominator gets replaced by the essentially constant host concentration. $\frac{LRT}{C_\mathrm{host}}$ is captured by a single parameter corresponding to the diffusivity $D$. This gives $\boldsymbol{J}=-D\nabla C$, with a mass balance giving $\dot C=D\nabla^2 C$, which are Fick's laws. To incorporate gravity, we'd go all the way back to the definition of the chemical potential and add an $mgz$ upward-movement energy penalty term to $G$, as before.
