For one dimensional non-relativistic quantum mechanics, the solutions to $\hat H\psi=E\psi$ seems not requiring the energy $E_n$ to contain the "$n$" term without specific boundary conditions.
Does the quantization come from that the wave function $\psi$ should vanish at infinity?
That is the boundary condition:
$$\psi\to0 \ \ \text{as} \ \ x\to\infty.$$
Sorry if the question is already asked.
I have no idea what you mean by "requiring a common n term" but it is very convenenient to think of quantization as resulting from imposing boundary conditions to solutions of the Schrodinger equation. This perspective is often found in elementary texts that present quantization by analogy to the appearance of (discrete) normal modes on a string or in a tube resulting from forcing a node or an antinode at one end (or both ends) of the physical system.
Specifically, it is not hard to find solutions of the Schrodinger equation for arbitrary potentials, but it is a lot more delicate to find solutions that satisfy the boundary conditions for this potential. This last requirement very severely restricts the possible choices of the eigenvalues: you can try increasing by a tiny amount the energy $E_n$ of the $n$'th eigenstate of the simple harmonic oscillator and the wave function will eventually diverge.
In the case of exactly solvable potentials, like the harmonic oscillator or the hydrogen atom, the boundary conditions at $\infty$ dictate the asymptotic behavior of $\psi(x)$ or $\psi(r)$. This asymptotic behavior is found as a first step and then included in an ansatz function that interpolates from $\infty$ back to finite values and must satisfy some differential equation for finite values of $x$ or $r$. Quantization "occurs" when one must select those interpolating functions that do not "undo" the asymptotic behavior already found.
If you can access it, this older paper by Sir Nevill Mott contains an excellent summary of the relation between quantization and boundary conditions.
(I see @tparker has already adressed the issue of compact support.)
You are correct that typically only bound states (corresponding to the boundary condition that your wrote down) have discrete energy levels.
But you used the term "compact support" the wrong way. "Compact support" doesn't mean the boundary condition that you wrote down; it means that the wavefunction is identically zero outside of some finite maximum radius $R$ from the origin, not just that it approaches zero. It is very rare for energy eigenstates to have compact support.
