fluid dynamics - Viscous drag proportional to $r$ or $r^3$?

When a spherical object is falling at terminal velocity through a fluid: $W=U+F$, where $W$ is weight, $U$ is upthrust and $F$ is viscous drag. Rewriting, using Stoke's Law, we get: $$\frac{4}{3}\pi r^3\rho_{object}.g=\frac{4}{3}\pi r^3\rho_{fluid}.g+6\pi r\eta v_t$$ $$6\eta v_t= \frac{4}{3} r^2\rho_{object}.g - \frac{4}{3} r^2\rho_{fluid}.g$$ $$v_t = \frac{2r^2(\rho_{object}-\rho_{fluid}).g}{9\eta},$$

where $r$ is the radius of the object, $\rho$ is density, $g$ is the gravitational acceleration, $\eta$ is the viscosity coefficient, and $v_t$ is the terminal velocity of the object.

So $v_t$ varies proportionally to $r^2$.

However, clearly upthrust varies proportionally to $r^3$. This implies that $F$ varies proportionally to $r$ so that $v_t$ can be proportional to $r^2$. But $F=6\pi r \eta v_t$, so as we increase $r$, $v_t$ increases quadratically, meaning that $F$ must increase cubically.

This seems to be a contradiction - $v_t$ is proportional to $r^2$ so $F$ is proportional to $r$, but also $F$ is proportional to $r^3$.

So where is my reasoning wrong, and does $F$ vary proportionally to $r$ or $r^3$?

EDIT:

I found this quote on the Wikipedia page for Stokes' law:

Note that since buoyant force increases as $R^3$ and Stokes drag increases as $R$, the terminal velocity increases as $R^2$ and thus varies greatly with particle size as shown below.