I would like to add a point to Stephan's answer:
White titanium dioxide is made up wavelength-scale particles, which scatter the light in a relatively wavelength-independent (Mie-type) manner. This scattering does rely on the refractive index being larger than one. Case in point, if the particles had $n=1$, they would be invisible (in air). This also relies on the TiO$_2$ refractive index having a very small imaginary part (absorption would make the film black or otherwise colored).
So while the large index of titanium dioxide does not make it "whiter" than other non-absorbing nanoparticles (in the sense of having a flatter scattering spectrum), you likely need a thinner layer of TiO$_2$ compared to other, lower index, materials to get the same scattering amplitude (opacity).
Update regarding comment:
There was a question regarding Mie scattering for dielectric nanoparticles. It turns out that Mie theory is often the framework of choice to describe light scattering from wavelength-scale particles, and it makes the physicist's favorite approximation: treating the particles like spheres. These particles can have any permittivity (i.e. dielectric, conducting, etc) as long as they have ~spherical symmetry, since the theory is really more about the spherical expansion of scattered waves than anything else.
As an example, I found a non-paywall paper in which the authors use Mie theory to compute scattering coefficients for TiO$_2$ nanoparticles with the aim of designing white surfaces.