This is a fantastic question! Let's get started.
I will assume that we already have defined the ladder operators $a$ and $a^{\dagger}$ and have defined a "ground state" $|0\rangle$ (we still have not proved it is the ground state) such that $a|0\rangle=0$. We will also assume that we already know that the Hamiltonian of the Harmonic oscillator can be written in the form
$$H=\hbar\omega\left(a^{\dagger}a+\frac{1}{2}\right).$$
(Note that the ground state $|0\rangle$ is trivially an eigenstate with $E_0=\hbar\omega/2$.) Finally, I will assume that we have already shown the commutation relations of the ladder operators. Namely,
$$[a,a^{\dagger}]=1.$$
With this, we have enough for a proof.
We can define a state $|n\rangle$ (let's forget about normalization for now) as
$$|n\rangle=(a^{\dagger})^n|0\rangle,$$
where $n$ is a nonnegative integer. The state $|n\rangle$ is an eigenstate of the Hamiltonian with energy $E_n=\hbar\omega(n+1/2)$. We wish to show that the set $\{|n\rangle\}_{n\in\mathbb{Z}^+}$ are all of the possible normalizable eigenstates of the Hamiltonian.
Recall that in the position representation, if we have a potential $V(x)$, then we cannot have a normalizable eigenstate $|\psi\rangle$ whose energy satisfies $E_{\psi}\leq\min V(x)$. That is, we can not have an energy less than the minimum potential energy of the system (ie the kinetic energy must be positive).
Now, we finish off with a proof by contradiction. Consider an eigenstate $|\psi\rangle$ whose energy is given by $E_{\psi}=\hbar\omega(n+1/2+\epsilon)$, with $\epsilon\in(0,1)$. Such a state would essentially describe any of the "other" states that $H$ could permit. Now, consider the state $|\psi^{(1)}\rangle=a|\psi\rangle$. By the commutator algebra, it is not hard to show that $|\psi^{(1)}\rangle$ has energy
$$E_{\psi^{(1)}}=\hbar\omega\left((n-1)+\frac{1}{2}+\epsilon\right).$$
Now, we can induct and define a state $|\psi^{(m)}\rangle\equiv(a^m)|\psi\rangle$. Clearly, its energy is given by
$$E_{\psi^{(m)}}=\hbar\omega\left((n-m)+\frac{1}{2}+\epsilon\right).$$
Thus, unless this process terminates at some point (that is, $a|\psi^{(m)}\rangle=0$ for some $m$), we can achieve an arbitrarily low energy. However, this process could never terminate, since the ground state $|0\rangle$ is unique (it's defined in terms of a position operator and a single derivative operator, so $a|0\rangle=0$ simply defines a first order differential equation in position space) and has energy $\hbar\omega/2$, this cannot be achieved for any $\epsilon$ in the given range. Thus, no such state $|\psi\rangle$ can occur. Similarly, we cannot have a state with energy $E_{\psi}\in(0,\hbar\omega/2)$ by the same logic.
Thus, we have (very rigorously) shown that the only normalizable of $H$ are those with energy $\hbar\omega(n+1/2)$, which are uniquely made from the action of ladder operators on the ground state.
I hope this helped!
(TL;DR -- If another state did exist, it would have an energy not of the form of those given by ladder operators. However, acting on this state many times with $a$ would produce an arbitrarily low energy, and is thus such a state could not exist.)
