Question: I know that it sounds like a very stupid question but if Shor's algorithm has a complexity of roughly $n^3$ why cant we just increase the bit size until the time for the algorithm to run is unfeasible on a quantum computer or would it just take too much memory and too much computation for RSA/ECC to be worth it?

Answer: 
The main reason is because any such technique has at best a *polynomial gap* between

1. the effort honest parties must spend to compute the cryptosystem, and
2. the effort adversaries must spend to break the cryptosystem.

Cryptography from a polynomial honest-to-malicious hardness gap has been known for a while.
In fact, one of the earliest public-key cryptosystems (*before* RSA by a few years iirc) went by the name of *Merkle's Puzzles*, and had precisely this property, namely they took $O(n)$ time to compute for honest parties, and $\Omega(n^2)$ for adversaries. See [Public-Key Cryptography in the Fine-Grained Setting](https://eprint.iacr.org/2019/625.pdf) for example.
Moreover, in the random oracle model, this is size of gap is [known to be optimal](https://www.iacr.org/archive/crypto2009/56770369/56770369.pdf), at least in the classical setting.

This is to say that if you are going to "settle" for a small gap between honest parties and adversaries, it is perhaps better to do other things than simply RSA "scaled up" appropriately, though perhaps it is best to simply use a lattice-based scheme instead.
That's what CNSA 1.0 did: minimum RSA key length is 3072, minimum AES key length is 256, and minimum SHA length is 384. [CNSA 2.0](https://media.defense.gov/2022/Sep/07/2003071834/-1/-1/0/CSA_CNSA_2.0_ALGORITHMS_.PDF) goes beyond that with some new algorithms.