Could the discovery of a counterexample to the Unit Conjecture change mathematicians’ understanding of spinors in general? Recently it has been reported that Giles Gardham has found a counterexample to the Unit Conjecture for group rings, as given in https://arxiv.org/abs/2102.11818 (and also in a popular article https://www.quantamagazine.org/mathematician-disproves-group-algebra-unit-conjecture-20210412/ )
Being from a chemistry background with an interest in spinors (not myself a mathematician), I have been advised that a spinor can be defined, from a mathematician's viewpoint, as "a unit length element of a division algebra". (I do know what division algebras are, but am not a competent algebraist.)
Assuming division algebras can contain group rings, then could the discovery of a counterexample to the Unit Conjecture lead to significant changes in the way mathematicians view spinors in general?
 A: Short answer: no.
I will split my answer into two parts:
Division algebras and group rings.
When it comes to spinors, the division algebras that are relevant are the real normed division algebras, namely $\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}$ (reals, complex numbers, quaternions, octonions). The octonions $\mathbb{O}$ are not quite relevant in the same way as $\mathbb{R},\mathbb{C},\mathbb{H}$ are since they are nonassociative, but they are still relevant. As real vector spaces these algebras have dimensions $1,2,4,8$.
If $G$ is a group and $R$ a subring of $\mathbb{C}$ with fraction field $K$, then the group ring $R[G]$ is a subring of the group algebra $K[G]$, and $K[G]$ is $|G|$-dimensional as a $K$-vector space. The structure theory of $K[G]$ for finite $G$, as an algebra, is described by the Artin-Wedderburn theorem: it is equivalent to a direct sum of matrix algebras over division algebras. The structure of group rings, despite simply being subsets of group algebras, are much harder to figure out however. Much can happen and much is open.
The unit conjecture pertains to when $G$ is torsionfree, however. That implies $G$ is infinite, so $K[G]$ is infinite-dimensional, and the usual Artin-Wedderburn theorem does not apply. Moreover, the counterexample to the conjecture uses $K=\mathbb{F}_2$ which is not a subfield of $\mathbb{C}$.
What are spinors?

a unit length element of a division algebra

Eh. I mean, I guess you could argue that sort-of works for a couple kinds of spinor. A succinct, suggestive epithet is: "spinors are the square root of geometry" (Atiyah, probably).
Consider an object with orientation entanglement. Wiggling the imaginary bands that entangle the object with its environment while keeping the object itself fixed doesn't count as changing the entangled-object system. If you spin the object $360^\circ$ the object is not back to normal, but if you spin it $720^\circ$ it is back to normal. This is illustrated by the so-called plate / belt / string / cup trick attributed to Dirac. (There are many versions of this trick, hence many object-based names.) What this means is that the rotation group $\mathrm{SO}(3)$, as a topological space in itself, has a noncontractible loop (the rotations around a given axis) which is unique up to homotopy (i.e. "wiggling"); traversing this loop twice (like a rubber band that is twice-wound) actually is contractible though! In other words, $\pi_1(\mathrm{SO}(3))=\mathbb{Z}_2$.
Thus, there are twice as many "spins" as there are "rotations." The act of forgetting the entanglement turns these "spins" into "rotations," which is a $2$-to-$1$ homomorphism $\mathrm{Spin}(3)\to\mathrm{SO}(3)$ from the group of 3D "spins" to the group of 3D rotations. (This usage of the term "spin" is not standard, but I like it.)
We can generalize this situation to $n\ge3$ dimensions. (Or even $p$ space and $q$ time dimensions, with pseudo-Euclidean spaces and not-positive-definite quadratic forms replacing the inner product.)
Even though $\mathrm{Spin}(n)$ acts (non-faithfully) on $\mathbb{R}^n$ by rotations, it can act on other vector spaces too. This is known as a representation. Compare with group actions: even if a group $G$ carries a natural action on an object $X$, its elements may also be reinterpreted as transformations of another object $Y$. For instance, if $G$ is the symmetry group of the cube, it acts on the set $V$ of $8$ vertices, $E$ of $12$ edges, $F$ of $6$ faces, $S$ of $4$ space diagonals, $A$ of $3$ axes, or $T$ of $2$ inscribed tetrahedra, etc. There are certain representations of $\mathrm{Spin}(n)$ whose elements (that is, the vectors being acted on) are called spinors.
The kernel of $\mathrm{Spin}(n)\to\mathrm{SO}(n)$ is the $360^\circ$ spin (around any axis, they're all equivalent), which you can call $-1$. The semicircle arc (of a one-parameter subgroup) from $+1$ to $-1$ in $\mathrm{Spin}(n)$, when applied to a vector in $\mathbb{R}^n$, rotates that vector all the way around back to itself. However, that same arc when applied to a spinor $\xi$ traces out an arc between $\pm\xi$. Thus, spinors are in some sense the "square roots" of vectors, as spins are the "square roots" of rotations.
Defining spinors formally and explicitly requires some work. First, suppose we learn how quaternions describe 3D rotations using conjugation. In view of the Cartan-Dieudonne theorem (rotations are products of evenly-many reflections), we can use unit vectors to define reflections (across perpendicular hyperplanes). Algebraically, this means we construct the Clifford algebra on $\mathbb{R}^n$. Establishing some recursive relations and base cases, we can show all these algebras are equivalent to matrix algebras over (associative) division algebras $\mathbb{R},\mathbb{C},\mathbb{H}$. These carry "standard" modules which can be used to define spinors, and $\mathrm{Spin}(n)$ is the group generated by products of evenly-many unit vectors within the Clifford algebras. (Technically, we identify the even subalgebra of a Clifford algebra with a smaller Clifford algebra.)
If you want more details, you'll have to read an actual introduction!
