The observation that 28 = 27 + 1 shows that it is possible to have consecutive perfect powers and perfect numbers . However, this must be very rare. Is it unique ?
Questions: $\;$1) Are there any other examples of this which are known? $\;$$\,$2) Is there a rigorous or heuristic proof that there are only a finite number of instances of consecutive perfect powers and perfect numbers ?
Thanks
Not a complete answer.
No odd perfect numbers are currently known, so I will focus on even perfect numbers.
An even perfect number $M$ is given by $$M = 2^{p-1}\left(2^p - 1\right)$$ where $2^p - 1$ (and therefore $p$) is prime.
We compute that: $$M + 1 = 2^{2p-1} - 2^{p-1} + 1$$ $$M - 1 = 2^{2p-1} - 2^{p-1} - 1$$
For $p = 2$, we have $$M + 1 = 7$$ $$M - 1 = 5,$$ which are not perfect powers $q^k, r^s$ for primes $q, r$ and integers $k, s \geq 2$.
Hence we have $p$ an odd prime (since $2^p - 1$ is a Mersenne prime), which gives $$M + 1 \equiv 1 \pmod 4$$ $$M - 1 \equiv 3 \pmod 4,$$ from which we obtain $M \pm 1 \ne 2^t$, for $t \geq 1$.
Added June 21 2016
Since $2 \equiv -1 \pmod 3$, we also have $$M + 1 = 2^{2p-1} - 2^{p-1} + 1 \equiv -1 - 1 + 1 \equiv 2 \pmod 3$$ $$M - 1 = 2^{2p-1} - 2^{p-1} - 1 \equiv -1 - 1 - 1 \equiv 0 \pmod 3,$$ while $$M - 1 \equiv (-7)^{2p-1} - (-7)^{p-1} - 1 \pmod 9$$ which is equivalent to $$M - 1 \equiv -{7^{2p-1}} - 7^{p-1} - 1 \pmod 9.$$ Suppose that $M - 1 \equiv 0 \pmod 9$, so that $${7^{2p-1}} + 7^{p-1} = 7^{p-1}\left(7^p + 1\right) \equiv 8 \pmod 9.$$ This last congruence is satisfied, for example, by $p = 3$.
Nonetheless, we obtain that $M + 1 \ne 3^u$, for $u \geq 1$.