Does your critic dislike that the argument seems not applicable over general rings? But it is: if there's an isomorphism over some ring $R$ then we can descend to a finitely generated subring and pass to the quotient by a maximal ideal to get such an isomorphism over a finite field, and then count points.

Or maybe your critic would prefer to invoke the fact that any group variety structure on an affine space over a field $k$ is unipotent, which ${\rm{SL}}_n$ is not? Here is a Grothedieck-style proof of this fact about affine space, exactly in the same spirit as the preceding argument: if an affine space over $k$ has a non-unipotent $k$-group structure then by increasing $k$ to an algebraic closure it would (by virtue of being smooth, connected, and affine) contain a nontrivial $k$-split $k$-torus as closed $k$-subgroup. We can then once again descend this property to a subring of $k$ finitely generated over $\mathbf{Z}$ and specialize to a finite field and conclude by counting points: the size of affine space over $k$ of size $q$ is a power of $q$, whereas the nontrivial $k$-split $k$-torus subgroup forces the total number of $k$-points to be divisible by $q-1$, so we get a contradiction as long as $q \ne 2$, and we can certainly always arrange that by increasing the finite field before "counting" anyway.

(There is a more "direct" proof of this general fact about group structures on affine spaces in Springer's book on algebraic groups, but it is kind of complicated. The specialization trick sure makes it easier, at the cost of better algebro-geometric technique to work over a base that is not a field during the middle of that process.)