Why does voltage drop across a resistor?
electrons pile up on 1 side of the resistor

That is entirely correct. And there is a &quot;shortage&quot; of electrons on the opposite side.

as far as I know, that's not the case.

It is the case.
Except in the case of a super-conductor, electrons interact with atoms which compose the conductor in a way that transfers energy and momentum to those atoms, and gives a random motion the electron. To use a simple model, an electron collides with an atom, and ricochets off in a random direction. The difference between the electron's initial momentum and final momentum is transferred to the atom, and the difference between the electron's initial energy and final energy is also transferred to the atom. While energy and momentum are conserved by such an event in our model, the flow of current is not. The &quot;forward&quot; motion of the charged electron, is replaced by the &quot;forward&quot; motion of a neutral atom. (Of course in a very short time, the atom will collide with another atom, but we will not concern ourselves with this at the moment). So these collections in our model create two effects. One they randomize motion, turning the organized motion of a current into heat. Second, they stop the flow of charge. If there were no electric field to start accelerate the original or a new electron, current would come to a stop there.
But we know that in a circuit with no branches and with an applied electromotive force current doesn't just stop at one place, but is uniform throughout the circuit? How does that happen? If the electrons flow did just stop when they collided with atoms, as in our model, then charge would pile up. But the piled up charge would create an electric field. This field would accelerate electrons away from it, on both sides of the pile-up. This would both reduce the current into the pile-up area, and increase the current leaving the pile-up area. Net accumulation stops when current into an area equals the net current leaving that area are equal. In equilibrium, the algebraic sum of all currents into a &quot;node&quot; (really anywhere) is 0. This is Kirchhoff's Current Law.
So where might electrons pile up? What is it's distribution? For this we need Maxwell's/Heaviside's Equations. (The equations we actually use today were first formulated by Heaviside based upon work done by Maxwell). The first equation we will use is
$$\nabla \times E = \frac{\partial B}{\partial t}$$
This says that the divergence of E is equal to the rate of change of the magnetic field with respect to time. We will assume that there is no time varying magnetic field in our model, so that the divergence of E is 0.
$$\nabla \times E = 0$$
Whenever a vector field has a divergence of 0, then that vector field is equal to the gradient of a scalar potential field. In our case, we will call that potential field \$V\$
$$E = \nabla V$$
\$V\$ is what we think of a the &quot;voltage&quot; at a point, when that voltage is well defined. \$E\$ is proportional to the force on a charge (in the case of no magnetic field), and is what causes a charge to accelerate.
In order for there to be a uniform current in a section of a resistive conductor, there needs to be a constant \$E\$ throughout that section. But then we turn to another Maxwell/Heaviside Equation
$$\nabla \cdot E = \frac{\rho}{\epsilon_0}$$
\$\rho\$ in this equation is net charge density, and \$\epsilon_0\$ is just a constant. This equation tells us then wherever there is a uniform E field (\$\nabla\cdot E =0\$) there is no net charge density. In other words, in a conductor with a uniform E field, there is no net charge density. Charges only build up at points where the E field changes.
But we already saw that in a resistive conductor with a uniform linear resistance, there is a constant E field, so in such a section of a resistive conductor, there is no charge accumulation. All the charge accumulation occurs at the interfaces where linear resistance changes.

My assumption is that voltage is dropping all over the circuit

That is also true. But remember that \$E = \nabla V\$. The voltage can drop all along the circuit, but \$E\$ remains uniform as long as the rate of voltage drop per distance is uniform.