Statement. Among the even numbers, only powers of two $2^x$ (with $x$ being a nonnegative integer) can be 2-superperfect numbers, and then if and only if $2^{x + 1} - 1$ is a Mersenne prime. (The default multiplier $m = 2$ is tacitly assumed from this point forward).

Proof. The only divisors of $n = 2^x$ are smaller powers of 2 and itself, $1, 2, \ldots , 2^{x - 1}, 2^x$ Therefore, the first iteration of the sum of divisors function is $$\sigma(n) = \sum_{i = 0}^x 2^i = 2^{x + 1} - 1 = 2n - 1.$$ If $2n - 1$ is prime, that means its only other divisor is 1, and thus for the second iteration $\sigma(2n - 1) = 2n$ and is thus a 2-superperfect number. But if $2n - 1$ is composite then it is clear that $\sigma(2n - 1) > 2n$ by at least 2. So, for example, $\sigma(8) = 15$ and $\sigma^2(8) = 24$ so 8 is not 2-superperfect. One more example: $\sigma(16) = 31$ and since 31 is prime, $\sigma^2(16) = 32$

Now it only remains to prove that no other even number $n$ can be 2-superperfect. Any other even number can of course still be divisible by one or more powers of two, but it also must be divisible by some odd prime $p > 2$ Since the sum of divisors function is a multiplicative function, it follows that if $n = 2^xp$ then $\sigma(n) = \sigma(2^x)\sigma(p)$ So, if, say, $p = 3$ it is clear that $(2^{x + 3} - 4) > 2^{x + 1}3$ and that on the second iteration this value that already exceeded twice the original value will be even greater. For example, $12 = 2^2 3$ and $\sigma(12) = 2^5 - 4$ which is greater than $2^3 3$ by 4. With any larger $p$ the excess will be much greater.