In the Schrödinger treatment of the Hydrogen atom, the excited states are true eigenstates, that is, their lifetime is infinite. This means that, if the Hamiltonian is
$$
H=\frac{P^2}{2m}+\frac{\alpha}{r}
$$
then $\tau=\infty$ for all the energy levels. What this description misses is the interaction of electrons with the electromagnetic field, that is, there is a
$$
j\cdot A
$$
term missing in $H$. Without this term, the electron doesn't couple to photons, and therefore the atom cannot emit them and decay. If you include the coupling to the electromagnetic field, for example with perturbation theory, you'll see that $\tau^{-1}\neq 0$ but it is very small (which nicely agrees with experiments by the way).
Finally, the most correct treatment of the Hydrogen atom is with QFT, that is, the Dirac+Maxwell equations (for fields instead of wavefunctions). In this case, you can reproduce the Schrödinger results, with relativistic corrections and fine structure and check that only the ground state is a true eigenstate (and the excited states are quasi-bound states, i.e., they have a small imaginary energy, that is, a finite life-time).
For more details, see this or this.
By the way, $\Delta E\sim\hbar/\tau$ has nothing to do with the uncertainty principle; see e.g. What is $\Delta t$ in the time-energy uncertainty principle?.
