To prove the existence of the ground state, I will follow the derivation made in Richard Feynman "Statistical Mechanics: A Set Of Lectures", Chapter 6, "A Simple Mathematical Problem".
Let us consider some (normalized) eigenstate $|\nu\rangle$ of the number operator $\hat{a}^\dagger \hat{a}$ with eigenvalue $\nu$. From commutation relations, we find that the states $|\psi_{\nu,k}\rangle =\hat{a}^k |\nu\rangle$ have eigenvalue $\nu - k$ for all $k$. Also, the norm of these states is
\begin{equation}
\langle\psi_{\nu,k}|\psi_{\nu,k}\rangle = \langle \nu | \hat{a}^k (\hat{a}^\dagger)^k|\nu\rangle = \nu(\nu-1)\dots(\nu - k + 1).
\end{equation}
From this equation, it directly follows that $\nu$ should be non-negative integer. Otherwise, there exist states $|\psi_{\nu,k}\rangle$ (with $k > \nu$) with negative squared norm in Hilbert space.
Therefore, the state $|\psi_{\nu,\nu} = \hat{a}^\nu|\nu\rangle$ is a ground state: it is an eigenvector of $\hat{a}^\dagger\hat{a}$ with eigenvalue $0$, and it is annihilated by $\hat{a}$.
The uniqueness of the vacuum state can't be proven just from the commutation relations.
Consider a direct sum of two independent bosonic or fermionic subspaces spanned by the states $|n\rangle_{1,2}$, where the subscript index indicates the subspace, and $n$ is the number of quanta. Assume that the creation/annihilation operator acts on each of the subspaces independently. Then, there are two vacuum states in the system. An example of such a system is a spin-$1/2$ particle in a harmonic potential: both of the states $|0, \uparrow\rangle$ and $|0, \downarrow\rangle$ are annihilated by $\hat{a}$.
