Let $\sigma(n)$ represent the divisor sum of $n$ and $\sigma^m(n)$ represent the repeated application of the divisor function $m$ times.
$(m,k)$-perfect numbers are numbers such that $\sigma^m(n) = kn$. For example, $\sigma^3(12) = 120 = 12\times10$, so $12$ is a $(3,10)$-perfect number.
It's easy to see that $\sigma(n) \ge n, \forall n \in \mathbb N$, and it follows that $\sigma^{m+1}(n) \ge \sigma^m(n), \forall n,m \in \mathbb N$.
Now, let $f(m, k)$ represent the smallest $(m,k)$-perfect number, i.e. the minimal solution to $\sigma^m(n) = kn$. For all $m,k$ that actually have a solution, is it possible to show that $f(m, k) \ge k, \forall k \in \mathbb N$?
No. If we think about $n$ as being fixed and allowing $m$ and (particularly $k$) to vary, we have that $n$ is an $(m,k)$-perfect number for some $k$ given any $n$ for which $$n\ \big|\ \sigma^m(n).$$ In particular, $$\sigma^4(2)=8,$$ so $2$ is $(4,4)$-perfect. In fact, it seems likely that the sequence $x_i=\sigma^i(2)$ contains infinitely many terms, which would imply that $2$ is $(m,k)$ perfect for some $k$ above any given threshold.