Based on some of your comments, I think what might be tripping you up is the first statement you started with:
From the Bohr's atomic model, it is clear that electron can have only certain definite energy levels.
and
...If suppose, we assume electron losses total energy, electron can't stay in any particular shell, as it would not have that particular value of energy.
That may be true for Bohr's atomic model, but Bohr's atomic model is wrong. And electron does not have to be in a particular, definite shell or energy level. Rather, any electron state is a superposition of states of definite energy level (energy eigenstates).
That means the expectation value of a hydrogen electron state is going to look like
$$\langle E\rangle = \sum_n |a_n|^2 E_n\text{,}$$
where $\{a_n\}$ are arbitrary complex values with $\sum_{n>0}|a_n|^2 = 1$ and $E_n$ are energy levels in increasing order. Because of the sum-to-$1$ condition, taking any portion along the other energy eigenstates will increase energy compared to the ground state.
In other words, even if the electron state does not have a definite energy, you still can't go lower than the ground state.
Suppose, I have a cup of hot coffee on the table. It will be continuously losing energy in the form of heat, but it stays on the table, though there was a energy loss. Now, all of a sudden, I take off the table, the cup of coffee converts it potential energy into kinetic energy to come down.
If you don't shake the table, the coffee cup will sit there, forever. Similarly, nothing perturbs the electron in an excited energy eigenstate, then it simply will never decay. It cannot: energy eigenstates are stationary; they do not evolve into anything other than themselves.
However, being completely without external perturbation is actually impossible. The uncertainty principle provides the electromagnetic field with vacuum fluctuations, which will perturb the electron even if nothing else in the environment does. In your analogy, this (or something else) provides the "shaking of the table" for the electron. Once the electron state gains even a tiny component in some other energy eigenstate, the state can evolve in time.
In other words, one can think of spontaneous emission as a particular type of stimulated emission where it's the vacuum that does the perturbing.
Best Answer
First you say
By way of preparation, I'll note that in introductory course work you never attempt to handle the multi-electron atom in detail. The reason is the complexity of the problem: the inter-electron effects (screening and so on) mean that it is not simple to describe the levels of a non-hydrogen-like atom. The complex spectra of higher Z atoms attest to this.
Later you say
but the best models of the nucleus that we have (shell models) do have nucleons occupying discrete orbital states in the combined field of the all the other nucleons (and the mesons that act as the carriers of the "long-range" effective strong force).
This problem is still harder than that of the non-hydrogen-like atoms because there is no heavy, highly-charged nucleus to set the basic landscape on which the players dance, but it is computationally tractable in some cases.
See my answer to "What is an intuitive picture of the motion of nucleons?" for some experimental data exhibiting (in energy space) the shell structure of the protons in the carbon nucleus. In that image you will, however, notice the very large degree of overlap between the s- and p-shell distributions. That is different than what you see in atomic orbitals because the size of the nucleons is comparable to the range of the nuclear strong force.