Let rings be unital but not necessarily commutative.
Suppose I have a ring $R$. Let a subring $k \subset R$ be called a subfield precisely when $k$ is a field, i.e. $k$ has distinct 0 and 1, all elements of $k$ except $0$ have multiplicative inverses and multiplication commutes when restricted to $k$.
Our ring $R$ has a lattice of subrings. If we take just the subfields we can get a poset, at least.
If we take $\mathbb{Z}$, for example, the subfield poset is empty.
If we take $\mathbb{Q}$, the subset poset is a singleton. Once we have $1$, we can get every rational number.
For $\mathbb{Q}[\sqrt{2}]$, we have $\mathbb{Q}$ and $\mathbb{Q}[\sqrt{2}]$.
However, for the quaternions with rational coordinates when thought of as a real vector space $\mathbb{Q}[i, j, k]$, our poset is not a lattice. $\mathbb{Q}[i]$, $\mathbb{Q}[j]$ and $\mathbb{Q}[k]$ are isomorphic fields, but a subring containing two of them will fail to be commutative because the whole ring is noncommutative.
For $\mathbb{Z}/p\mathbb{Z}$, once we have $1$ we have the whole ring.
For $\text{GF}(p^2)$, I think we have the field containing just $1$, which is isomorphic to $\text{GF}(p)$, and the whole field, which is isomorphic to $\text{GF}(p^2)$.
Is there a way to characterize which posets we can hit in general in this manner?