Why can't the energy of an electron exceed 0 eV?
The value of the energy in the Bohr model is zero when the quantum number is infinity  because that is the limiting value of the Coulombic potential at large distances, and because the electron is assumed bound to the nucleus (the atom is stable), which constrains the value of the total energy.
The energy of a stationary hydrogen-like atom is described by the equation
$$ U(Z,n) = R_\mathrm{E} \frac{Z^2}{n^2} $$
where $R_\mathrm{E} = \pu{-13.6 eV}$. It can equally well be written as
$$ U(Z,n) = \frac{-Zk_e e^2}{2r_n} \tag{1}$$
where $r_n$ is the allowed orbit radius in the Bohr model, which illustrates how the energy becomes more positive (goes to zero) as the average distance between the nucleus and the electron increases.
The Bohr model assumes the universe consists of just one electron and nucleus. It is implicit in the model that the electron is bound (the electron and nucleus can never get far enough apart to escape their mutual attraction). According to the virial theorem the total energy in the above equations is one half of the Coulombic potential energy and the negative of the kinetic energy. That means the expectation values of the potential and kinetic energy for a given quantum number are fixed, and that the value of the total energy is bounded by the value of the potential energy (see Eq. 1):
$$U(Z,n)=\frac{\left&lt;U_{\mathrm{Coulomb}}\right&gt;}{2}$$
The zero limiting value of the Coulombic potential can be interpreted as indicating that the interaction between the particles vanishes at large distances. On the other hand, the zero limiting value of the total energy can be interpreted as indicating that the reference frame of the atom is at rest. If the atom is assumed to have kinetic energy than this can be represented as a non-zero  offset of the total energy. In that case the energy converges to this threshold as the particles move farther apart.
In addition, in practice external fields might interact with the atom, contributing to the potential. It is possible for instance to superimpose a uniform linear potential on the Coulombic potential. The solutions (wavefunctions) for the new perturbed atom are not identical to those of the original isolated atom. Suffice it to say that if you now pull the particles apart far enough  they will eventually be free to follow trajectories that are no longer bound.
The following figures illustrate (roughly) the energies of an atom (for Z=1) before and after superimposing an external linear potential, plotted against the quantum number or the orbit radius in the original Bohr model. In the case of particles starting at rest, what determines whether particles will remain bound or fall apart is relative values of the potential. It is the slope of the potential that determines forces that drive particles together or apart. In the last figure, the (slight) downward slope at large r as the radius increases indicates that particles at rest starting at this internuclear distance are no longer bound.
Hydrogenic atom in Coulombic potential...


... and perturbed by linear potential

An electron can also be excited by a photon into the continuum of non-quantized states with energy greater than zero, but then the energy is no longer described by the Bohr model. The excess energy (greater than the ionization energy) amounts to the kinetic energy of the dissociated (nonbound) state (which is alluded to in the other answer).
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