This immediately follows from the fact that $\frac{\sigma(m)}m$ is unbounded (any unbounded from above sequence has infinitely many terms dominating all previous terms). To show the latter, just look at the product of the first several primes $m=p_1\dots p_n$, note that $\frac{\sigma(m)}m=\prod_{j=1}^n (1+p_j^{-1})\ge\sum_{j=1}^n p_j^{-1}$ and recall that the series of inverse primes diverges.

Let $f(n)=\frac{\sigma(n)}{n}.$ Claim: Let $p_i$ denote the $i$th prime. Then, the product $$\prod_{i=1}^\infty (1+\frac{1}{p_i})$$diverges. Clearly, we have $$\prod_{i=1}^\infty (1+\frac{1}{p_i})\geq 1+\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}\cdots.$$It is well known that the sum of the reciprocals of the primes diverges, so this shows the claim. Since $\sigma$ and thus $f$ is multiplicative, we have $$f(p_1)=(1+\frac{1}{p_1}),f(p_1p_2)=(1+\frac{1}{p_1})(1+\frac{1}{p_2}),f(p_1p_2p_3)=(1+\frac{1}{p_1})(1+\frac{1}{p_2})(1+\frac{1}{p_3}),$$and so on. Since this product diverges, the $f$ function can get arbitrarily large. Since it can get arbitrarily large, it must hit a new maximum infinitely often, QED.

does $2^n$ work? we get $\frac{\sigma(2^n)}{2^n} = 2-\frac{1}{2^n}$ and seems like we are able to prove that $2^n$ works for every natural $n$

Bataw wrote: does $2^n$ work? we get $\frac{\sigma(2^n)}{2^n} = 2-\frac{1}{2^n}$ and seems like we are able to prove that $2^n$ works for every natural $n$ What do you mean by "works"? Certainly, $2^n$ is never superabundant for any $n \ge 3$ (as $\frac{\sigma(6)}{6}=2$).