Yes, Coulomb’s law is only for point charges separated by a distance $r$. The inverse square law is there because of the diverging or converging nature of electric field from a point charge which is the crucial point. Here is how I explain:
Imagine spherical surfaces with increasing radius around a point charge. As we increase the radius, the intensity of electric field per unit area of the spherical surface goes on decreasing as $\frac{1}{r^2}$. This means the intensity at any point $r$ decreases as we increase $r$. This is because of the diverging or converging of the electric field of the point charge. An example to understand better: consider particles coming out of a point and they come out with finite numbers forming a spherical surface. As they come out, sphere becomes larger and larger but particles are finite on the surface. So, particle density on the surface of the sphere goes on decreasing. So,the number of particles at $r$ decreases as $r$ increases.
This is what Gauss law tells, when we integrate electric field intensity over the whole surface of larger radius, we should get the same value when we integrate electric field intensity over surface for smaller radius which is $\frac{q}{\epsilon_0}$ which is finite.
But, electric field from charged planes doesn’t diverge or converge in ideal case, it is always perpendicular to the plane since the plane is an equipotential surface. Since electric field doesn’t diverge or converge but are perpendicular to the plane, the electric field remains the same at any point near or far from the plane, which is $\frac{\sigma}{2\epsilon_0}$ from Gauss law. So, the electric field between two equally charged planes with $+Q$ remains always $0$ , $\frac{\sigma}{2\epsilon_0}-\frac{\sigma}{2\epsilon_0}=0=\frac{dV}{dr}$ or the potential at any point between them remains constant, the test charge would remain wherever it is placed!
