Combinatorial interpretation of Euclid's form for even perfect numbers

Euclid showed that if $p$ is a prime such that $2^{p}-1$ is also a prime, then the number $n=2^{p-1}.(2^{p}-1)$ is perfect. Much later, Euler proved that every even perfect number is of this form. Thinking of proper divisors of a number as analogues of proper subsets of a discrete set, one can see an analogy between this number of subsets ($2^{p}-1$ for a set of $p$ elements) and the Mersenne prime occuring in Euclid's formula.
So my question is: as a perfect number is defined as a number equal to the sum of its proper divisors, is there a combinatorial interpretation of Euclid's form explaining why an even perfect number is of the aforementionned form? Moreover, if $m$ is a (not necessarily even) perfect number, must there exist an integer $k$ such $2^{k}-1$ divides $m$?
Thanks in advance.

Comment converted to an answer, so that this question does not remain in the unanswered queue.

If $m$ is (even) perfect, then $m = 2^{p-1} (2^p - 1)$, for some Mersenne prime $2^p - 1$. So taking $k = p$, the answer to your second question is yes for even perfect numbers. As for odd perfect numbers $M$, it is currently unknown whether $2^2 - 1 = 3 \mid M$. As to your first question: I am currently unaware of any combinatorial interpretation of Euclid's form explaining why an even perfect number is of the aforementioned form.