The time-independent Schrödinger equation is
$$\dfrac{-\hbar^2}{2m} \nabla^2 \psi + V \psi = E \psi.$$
I am then told that the general solution for a uniform potential can be written as the sum of two counter-propagating plane waves,
$$\psi(r) = Ae^{ikz} + Be^{-ikz},$$
where
$$k^2 = \dfrac{2m}{\hbar^2}(E - V)$$
is found by substitution back into the time-independent Schrödinger equation.
I would greatly appreciate it if people would please take the time to show me how the general solution $\psi(r) = Ae^{ikz} + Be^{-ikz}$ was found. Furthermore, I would greatly appreciate it if people would explain what is meant by the general solution for a "uniform potential", and how the two plane waves are "counter-propagating".
My work:
$$\dfrac{-\hbar^2}{2m} \nabla \psi(r) + V \psi(r) = E \psi(r)$$
$$\begin{align} &\therefore \dfrac{-\hbar^2}{2m} \dfrac{d^2 \psi}{dr^2} + V \psi = E \\ &\Rightarrow \dfrac{\hbar^2}{2m} \dfrac{d^2 \psi}{dr^2} + (E - V) \psi = 0 \\ &\Rightarrow \dfrac{d^2 \psi}{dr^2} + \dfrac{2m}{\hbar^2}(E - V)\psi = 0 \end{align}$$
So we have the following characteristic equation:
$$\begin{align} &k^2 + \dfrac{2m}{\hbar^2}(E - V) = 0 \\ &\Rightarrow k^2 = \dfrac{2m}{\hbar^2}(V - E) \\ &\Rightarrow k = \pm \sqrt{\dfrac{2m}{\hbar^2}(V - E)} \end{align}$$
If we let $A = \sqrt{\dfrac{2m}{\hbar^2}(V - E)}$, then we have that
$$\begin{align} \psi(r) = C_1 e^{A} + C_2 e^{-A}, \\ \psi'(r) = C_1A e^{A} - C_2 A e^{-A}, \\ \psi''(r) = C_1 A^2 e^{A} + C_2 A^2 e^{-A} \end{align}$$
But this is not the same as $\psi(r) = Ae^{ikz} + Be^{-ikz}$. And how do we account for the fact that we have a "uniform potential"?
The Schrödinger equation describes the evolution of the wave function of a particle (or system). When you assume that the wave function $\Psi(x, t)$ is separable, i.e., $\Psi (x,t) = \psi (x) T(t)$, you obtain the following Schrödinger equation, $$-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x,\,t) + V(x)\Psi(x,t) = i\hbar\frac{\partial}{\partial t} \Psi(x,\,t).$$ This equation is equivalent to these two equations: \begin{align} & i\hbar \frac{d T(t)}{dt} = E \, T(t) \\ & -\frac{\hbar^2}{2m} \frac{d^2 \psi (x)}{dx^2}+V(x) \psi (x) = E \, \psi (x), \end{align} where the second equation is known as the time-independent Schrödinger equation.
The solution to the first equation is given by $$T(t) = e^{-i\frac{Et}{\hbar}}.$$ When you have a uniform potential, i.e. the function $V$ is constant, the second equation can also be solved easily using the method you described. The characteristic equation is given by: $$k^2 + \dfrac{2m}{\hbar^2}(E - V) = 0.$$ There are two cases, either $E-V < 0$. In that case the solution is given by: $$\Psi(x) = A \exp(-k x) + B \exp(kx), \quad k = \sqrt{\frac{2m}{\hbar^2}(V-E)}.$$ Either $E-V > 0$. In that case the solution is given by: $$\Psi(x) = A \exp(-ik x) + B \exp(ikx), \quad k = \sqrt{\frac{2m}{\hbar^2}(V-E)}.$$ That is for the mathematical part.
Now physically you have some constraints on your wave function. The first one being that, as $|\Psi(x)|^2$ represents a probability distribution, as such it must sum to one. The wave function is also continuous and differentiable. In a typical problem, you are given a potential and all these conditions gives you constraints on the possible solutions.
The canonical example is an infinite potential well, i.e. a potential that is finite between $0$ and $L$, and infinite everywhere else. In that case, it is easy to see that $\Psi(x)$ must be zero where the potential is infinite and so that by continuity of the wave function the wave vector $k$ can only take discrete value. Hence the quantization of the energy. See this article https://en.wikipedia.org/wiki/Particle_in_a_box
You can also study cases where the potential is piecewise constant. This gives rise to funny quantum phenomena such as the tunnel effect. See https://en.wikipedia.org/wiki/Finite_potential_well
Hope I answered your question.