It is well known that all even perfect numbers are of the form $N=(2^{q}-1).2^{q-1}$ with $M_{q}:=2^{q}-1$ a Mersenne prime.
As the very defining property of such a perfect number is to fulfill the equality $\sigma(N)=2N$, one can see that this value is almost the sum of a geometric series. But another conceptual framework can emerge from this: namely that $\sigma(N)$ is (close to) a $2$-adic series.
As Euler showed unrigorously, one has $\sum_{k=0}^{\infty}2^{k}=-1$, the latter being interpreted as convergence in $\mathbb{Z}_{2}$, the ring of $2$-adic integers. So a proof of the existence of infinitely many even perfect numbers should be morally equivalent to the convergence of the sequence $(2N)_{N}$ where $N$ runs over the even perfect numbers towards $0$ (edited after Wojowu's comment, I wrote $-1$ at first) in $\mathbb{Z}_{2}$.
My question is: can this be generalized to hypothetical odd perfect numbers? Namely, should $\sigma(N)$ for $N$ an OPN be "close" to some $p$-adic series for some $p>2$? Is there some hint that this should indeed be the case?
Edited after JoshuaZ' supposedly incomplete but insightful answer:
The number $2$ in the perfect number defining equality $\sigma(N)=2N$ may be viewed as the Euler factor at $p=2$ of $\zeta(s)$ as $s$ tends to $1$, that is, $N=\sigma(N)\lim_{s\to 1}\frac{\zeta_{2}(s)}{\zeta(s)}$ where $\zeta_{2}$ is the $2$-adic zeta function. Could we be able to prove that any perfect number $N$ whose smallest prime factor is $p$ fulfills $N=\sigma(N)\lim_{s\to 1}\frac{\zeta_{p}(s)}{\zeta(s)}$, this would imply $p=2$ and that no OPN exists.
Second edit: one can define the "numerical $p$-adic transform" of an L-function $F$ whose sequence of Dirichlet coefficients is $(a_{n}(F))_{n>0}$, so that $F(s)=\sum_{n>0}a_{n}(F).n^{-s}$ whenever $\Re(s)>1$, by $\mathcal{L}_{p}(F):=\sum_{n>0}a_{n}(F)p^{n}$ in $\mathbb{Z}_{p}$. That way perhaps a link between $\mathcal{L}_{2}(\zeta)$ and RH could be made.
Best Answer
(Not a complete answer but a bit too long for a comment.)
There's a fundamental difficulty here in proving the sort of result you envision. If there were some prime $p$ which had to divide every odd perfect number and larger odd perfect numbers had to be divisible by higher powers, that would work. But we can't right now rule out now even that for any given odd prime $p$, there are infinitely many odd perfect numbers with smallest prime factor $p$. As far as I'm aware, the closest thing we have to a restriction on that are the results in this recent paper of mine, especially Theorem 8 on page 47.
The closest thing in the literature might be some of what Tomohiro Yamada has done. One goal that Tomohiro Yamada has been working on has been to rule out certain families of exponents. One of his strongest results of this sort is the following: Let $N$ be an odd perfect number and write $$N= q^ap_1^{2e_1}p_2^{2e_2}\cdots p_{k-1}^{2e_{k-1}}$$ where $q \equiv a \equiv 1$ (mod 4) and $p_1$, $p_2 \cdots $ $p_{k-1}$ are distinct primes none equal to $q$. (Note that by Euler's theorem for odd perfect numbers we can always put one in this form.) Then if all the $e_i$ are the same. That is, $e_1=e_2...=e_{k-1}=e$, then we have $k \leq 2e^2 + 8e+3$. Note that since there are bounds on the size of an odd perfect number in terms of its number of distinct prime factors, this automatically says that there are only finitely many odd perfect numbers which all share the same exponent for all but one prime.
The obvious stronger goal to aim for is a theorem of the same flavor as above, but where instead of assuming that all the exponents are the same, construct a function $f(x)$ such that if all the $e_i$ are bounded above by some constant $E$, then $k \leq f(E)$. The results from Tomohiro use high powered sieve theory, but it seems like not many people have thought that hard about extending them, so I can't speak about how plausible this goal is given the current power of machinery. It might be the sort of thing that a group of very dedicated people might prove or it might be completely out of reach with current technology.
That said, if one did have a theorem of the sort envisioned in the last paragraph, one would then have as a corollary that for any $m$ there are only finitely many odd perfect numbers not divisible by a perfect $m$th power of a prime. So this would move sort of in the direction you want. But even this would be a very weak move in that direction because there's no guarantee that the primes are the same in each case.
Finally, let me add one piece of almost completely unfounded speculation, of how one might be able to get a result of the sort you want. It isn't hard to write down a Dirichlet series in terms of $\zeta(s)$ where the series has a non-zero $n$th term if and only if $n$ is an odd perfect number. Sketch of argument: write down the Dirichlet series for $(\sigma(n)-2n)^2$ using that $$\sum_{n=1}^{\infty} \frac{\sigma(n)^2}{n^s} = \frac{\zeta(s)\zeta(s-1)^2\zeta(s-2)}{\zeta(2s-2)}. $$ (The above is a special case of a general theorem for a Dirichlet series for $\sigma_a(n)\sigma_b(n)$. See e.g. Theorem 305 in Hardy and Wright.) Then remove all the even terms from the Dirichlet series, and you now have a series which has a term which is zero exactly where the odd perfect numbers are. One would hope that one could then maybe turn analytic methods to estimate the coefficients of this Dirichlet series. I've never been successful in getting anything non-trivial out of this, but it is possible that someone who is better at complex analysis than I am could make this work. But your question brings up a related idea had not occurred to me until now and which might be worth someone thinking about: how does the resulting series behave $p$-adically? Vague hand-wave to Kubota-Leopoldt goes here. If that could work it might be possible to go in the other direction, do some sort of $p$-adic approach to the Dirichlet series in question and see if that gives us new information about odd perfect numbers.