RSA using 2 as a public exponent I'm failing to see why 2 can never be used and what weaknesses would be associated with doing so. 

There's a similar question asking why it has to be in the form of $2^n$+1 but why not $2^n$
It is not that RSA becomes insecure when used with public exponent $2$. (In fact, the Rabin Cryptosystem does exactly that.) It's that it doesn't  actually work.

The problem is that for $N = pq$ for two primes $p$ and $q$, the function $f : x \mapsto x^2 \bmod N$ is not injective. So it is impossible to invert uniquely. In fact, most elements of the function's range (the set of perfect squares) have $4$ preimages under $f$. (As pointed out by Thomas Pornin, the multiples of $p$ and $q$ have $2$, while $0$ has only one.)

We can get around this problem by choosing $p\equiv q\equiv 3 \bmod 4$ and restricting the domain to the set of quadratic residues. In this case, the function is a trapdoor permutation over the set of quadratic residues if factoring is hard. 

One might note, that this is actually a better guarantee than the one we have for the RSA trapdoor permutation, where the security is not known to be implied by the hardness of factoring.