ag.algebraic geometry - What are the higher homotopy groups of Spec Z ?

$Spec(mathbb{Z})$ should only be considered as $S^3$, if you "compactify" that is add the point at the real place. This is demonstrated by taking cohomology with compcat support.
The étalé homotopy type of $Spec(mathbb{Z})$ is however contractible (indeed what do you get by removing a point form a sphere?) to see this (all results apper in Milne's Arithmetic Dualities Book)

(let $X=Spec(mathbb{Z})$ )

1) $H^r_c(X_{fl},mathbb{G}_m)=H^r_{c}(X_{et},mathbb{G}_m) = 0$ for $r neq 3$.

2) $H^3_c(X_{fl},mathbb{G}_m)=H^3_{c}(X_{et},mathbb{G}_m) = mathbb{Q}/mathbb{Z}$

3) by 2+1, we have:

$H^3_c(X_{fl},mu_n)= mathbb{Z}/n$

$H^r_c(X_{fl},mu_n)= 0$ for $r neq 3$.

4) since we have a duality $$H^r(X_{fl},mathbb{Z}/n)times H^{3-r}_c(X_{fl},mu_n) to mathbb{Q}/mathbb{Z} $$
we have

5) $H^0(X_{fl},mathbb{Z}/n) = H^0(X_{et},mathbb{Z}/n) = mathbb{Z}/n,$

$H^r(X_{fl},mathbb{Z}/n) = H^r(X_{et},mathbb{Z}/n) = 0$, $r >0$

6) Now since $pi_1$ is trivial we have by the Universal Coefficient Theorem, the Hurewicz Theorem and the profiniteness theorem for 'etale homotopy that all homotopy groups are zero.