Illusie in "Grothendieck et la cohomologie étale" says Artin's Harvard notes on *Grothendieck Topologies* prove: The étale cohomology with coefficients in $Z/nZ$ of the variety $\mathrm{Spec}(k[X,Y])\backslash\langle0,0\rangle$ for any algebraically closed field $k$ agrees with the cohomology of the 3-sphere. 


This seems to refer to 3.5 on Artin's p. 110. But that result is actually stated a little differently. It describes the étale cohomology with those coefficients of $\mathrm{Spec}(k\{X,Y\})\backslash\langle0,0\rangle$ where $k\{X,Y\}$ is the the Henselization of $k[X,Y]$ localized at the origin. Artin does suggest you should think of this scheme as a 4 ball minus a point. (Also Artin notes $k$ need only be separably algebraically closed.)


Does Artin's 3.5 somehow easily yield the result for $\mathrm{Spec}(k[X,Y])\backslash\langle0,0\rangle$?


Besides that Will Sawin's answer is easy enough for me to see, it is easy in a historic sense: These are largely considerations Artin would have known in the fall of 1961. He did not have the affine dimension theorem in general -- but he did for curves. If he did not yet have a rigorous étale Kunneth formula, he expected something like it, and might have worked around it in the case of constant sheaves on $\mathbb A^2$ over a separably closed field. So this supports the idea that this result on $\mathrm{H}^3$ of $\mathbb A^2 \backslash \langle 0,0\rangle$ is what Artin later called the first theorem in higher dimensional étale cohomology (in an interview, in Joel Segel, ed., *Recountings: Conversations with MIT Mathematician*s, A K Peters, 2009, pp. 351–74). And it seems this could be close to how he first proved it.



Etale cohomology of $\mathrm{Spec}(k\{X,Y\})\backslash\langle0,0\rangle$
I'm not sure what "easy" means in the context of etale cohomology but there is a way of passing from Artin's result to the stated one.


Let $j$ from $\mathbb A^2 \backslash \langle 0,0\rangle$ to to $\mathbb A^2$ be the open immersion. Then there is a Leray spectral sequence relating the etale cohomology of $\mathbb A^2 \backslash \langle 0,0\rangle$ to the etale cohomology of $\mathbb A^2$ with coefficients in $R^i j\_\* \mathbb Z/n\mathbb Z$. 


The key point is that the stalk of $R^i j\_\* \mathbb Z/n\mathbb Z$ at $\langle 0,0\rangle $ is precisely the $i$th etale cohomology group of $\operatorname{Spec} k\{X,Y\} \langle 0,0\rangle $ - because by definition it is the limit of the etale cohomology groups of punctured neighborhoods of $ \langle 0,0\rangle $, and these converge to the etale cohomology group of $\operatorname{Spec} k\{X,Y\} \langle 0,0\rangle $.


Furthermore, note that $R^0 j\_\* \mathbb Z/n\mathbb Z$ is the constant sheaf $\mathbb Z/n\mathbb Z$ by direct computation and $R^i j\_\* \mathbb Z/n\mathbb Z$ vanishes away from $\langle 0,0\rangle $ for $i>0$.


So $R^i j\_\* \mathbb Z/n\mathbb Z$ is zero for $i \not\in\{ 0,3\}$, is the constant sheaf $\mathbb Z/n\mathbb Z$ for $i=0$, and is the skyscraper sheaf for $i=3$. Taking cohomology, $H^p(\mathbb A^2, R^q j\_\* \mathbb Z/n\mathbb Z)$ vanishes for $p \neq 0$ or $q \not\in\{ 0,3\}$ and is $\mathbb Z/n$ for $p=0, q\in\{0,3\}$ so the spectral sequence degenerates and we get the desired result.


