In the formulation, presumably on the right side what is intended are 3-dimensional non-degenerate quadratic spaces (up to isomorphism), with discriminant 1 (same as $4^3$ mod squares as John Ma notes). But to make this work also in characteristic 2, it is better to proceed with a different point of view: that of conformal isometry of quadratic spaces (i.e., isomorphisms $T:V \simeq V'$ such that $q' \circ T = \lambda q$ for some $\lambda \in k^{\times}$). More specifically, we claim that away from characteristic 2, every 3-dimensional non-degenerate quadratic space is conformal to a unique one with discriminant 1. Thus, by working with conformal isometry classes we will be able to work in a fully characteristic-free manner.
To see what is going on, recall that the set of isomorphism classes of central simple algebras of dimension 4 is ${\rm{H}}^1(k, {\rm{PGL}}_2)$, and the set of conformal isometry classes of dimension 3 is ${\rm{H}}^1(k, {\rm{PGO}}_3)$. But ${\rm{GO}}_{2m+1} = {\rm{GL}}_1 \times {\rm{SO}}_{2m+1}$, so ${\rm{PGO}}_{2m+1} = {\rm{SO}}_{2m+1}$. Hence, ${\rm{PGO}}_3 = {\rm{SO}}_3$. Since ${\rm{SO}}_3 \simeq {\rm{PGL}}_2$ through the representation of ${\rm{PGL}}_2$ via conjugation on the 3-dimensional space of traceless $2 \times 2$ matrices equipped with the determinant as the standard split non-degenerate quadratic form $xy - z^2$ (preserved by that conjugation action!), that answers the entire question at the level of isomorphism classes of objects. (The link to ${\rm{SO}}_3$ encodes the link to discriminant 1.)
But we can do better than keep track of isomorphism classes: we can also keep track of isomorphisms, as explained below. This is a refinement of John Ma's answer, as well as that of Matthias Wendt (which appeared at almost exactly the same time as this answer first appeared, so I didn't see it until this one was done).
The following notation will permit considering finite fields on equal footing with all other fields. For a finite-dimensional central simple algebra $A$ over an arbitrary field $k$, let ${\rm{Trd}}:A \rightarrow k$ be its "reduced trace" and ${\rm{Nrd}}:A \rightarrow k$ be its "reduced norm". These are really most appropriately viewed as "polynomial maps" in the evident sense. That is, if $\underline{A}$ is the "ring scheme" over $k$ representing the functor $R \rightsquigarrow A \otimes_k R$ (i.e., an affine space over $k$ equipped with polynomial maps expressing the $k$-algebra structure relative to a choice of $k$-basis) then we have $k$-morphisms ${\rm{Trd}}:\underline{A} \rightarrow \mathbf{A}^1_k$ and ${\rm{Nrd}}:\underline{A} \rightarrow \mathbf{A}^1_k$.
For $A$ of dimension 4 we set $\underline{V}_A$ to be the kernel of ${\rm{Trd}}:\underline{A} \rightarrow \mathbf{A}^1_k$; speaking in terms of kernel of ${\rm{Trd}}$ is a bit nicer than speaking in terms of orthogonal complements so that one doesn't need to separately consider characteristic 2 (where the relationship between quadratic forms and symmetric bilinear forms breaks down). This $\underline{V}_A$ is an affine space of dimension 3 over $k$ on which ${\rm{Nrd}}$ is a non-degenerate quadratic form $q_A$ (i.e., zero locus is a smooth conic in the projective plane $\mathbf{P}(V_A^{\ast})$, where $V_A := \underline{V}_A(k)$): indeed, these assertions are "geometric" in nature, so it suffices to check them over $k_s$, where $A$ becomes a matrix algebra and we can verify everything by inspection.
We will show that the natural map of affine varieties
$$\underline{{\rm{Isom}}}(\underline{A}, \underline{A}') \simeq \underline{{\rm{CIsom}}}((\underline{V}_A, q_A), (\underline{V}_{A'}, q_{A'}))/\mathbf{G}_m$$
from the "isomorphism variety" to the "variety of conformal isometries mod unit-scaling" is an isomorphism; once that is shown, by Hilbert 90 we could pass to $k$-points to conclude that isomorphisms among such $A$'s correspond exactly to conformal isometries among such $(V_A, q_A)$'s up to unit scaling. It suffices to check this isomorphism assertion for varieties over $k_s$, where it becomes the assertion that the natural map
$$\underline{{\rm{Aut}}}_{{\rm{Mat}}_2/k} \rightarrow {\rm{CAut}}_{({\rm{Mat}}_2^{{\rm{Tr}}=0}, \det)/k}/\mathbf{G}_m$$
from the Aut-scheme to the scheme of conformal isometries up to unit-scaling is an isomorphism. But this is precisely the natural map
$${\rm{PGL}}_2 \rightarrow {\rm{PGO}}(-z^2-xy) = {\rm{PGO}}(xy+z^2) = {\rm{SO}}(xy+z^2) = {\rm{SO}}_3$$
between smooth affine $k$-groups that is classically known to be an isomorphism over any field $k$ (can check bijectivity on geometric points and the isomorphism property on tangent spaces at the identity points).
Finally, we want to show that every 3-dimensional non-degenerate quadratic space $(V, q)$ is conformal to $(V_A, q_A)$ for some $A$.
Note that if such an $A$ exists then it is unique up to unique isomorphism in the sense that if $A$ and $A$ are two such equipped with conformal isometries $(V_A, q_A) \simeq (V, q) \simeq (V_{A'}, q_{A'})$ then this composite conformal isometry arises from a unique isomorphism $A \simeq A'$ of $k$-algebras. Hence, by Galois descent (!) it suffices to check existence over $k_s$! But over a separably closed field the smooth projective conic has a rational point, so $(V, q)_{k_s}$ contains a hyperbolic plane and thus is isometric to $xy + \lambda z^2$ on $k_s^3$ for some $\lambda \in k_s^{\times}$. This is conformal to $(-1/\lambda)q_{k_s}$, but $(1/\lambda)xy - z^2$ and $-x'y' - z^2$ are clearly isometric, and the latter is ${\rm{Mat}}_2^{{\rm{Tr}}=0}$ equipped with the restriction of det.
