Among the first results one usually sees, right after defining schemes, is that affine schemes are compact, however this statement is stronger than $\mathsf{ZF}$: In $\mathsf{ZFC}$ we have that $$\operatorname{Spec}\left(\prod_{i=1}^\infty\Bbb F_2\right)\cong\beta\Bbb N,$$the Stone-Čech compactification on $\Bbb N$, but in $\mathsf{ZF}$ the only ultrafilters we can prove to exist on $\Bbb N$ are the principal ones, corresponding to $\Bbb N\subseteq\beta\Bbb N$, which is a discrete subspace, hence not compact. So, if $\beta\Bbb N\setminus\Bbb N=\varnothing$ we have an affine scheme which is not compact.
Is the statement "every affine scheme is compact" equivalent to some more well-known statement over $\mathsf{ZF}$? Or does it imply/is it implied by some more well known statements weaker than $\mathsf{AC}$?
The statement that every affine scheme is compact is equivalent to the Boolean prime ideal theorem, which is a well-known weak form of choice (strictly weaker than full AC) which is equivalent to many other common applications of choice (for instance, it is equivalent to the compactness theorem in first order logic, or Tychonoff's theorem for Hausdorff spaces).
Indeed, the usual proof that an affine scheme is compact uses the axiom of choice only to say that any nonzero commutative ring has a prime ideal, which is known to be equivalent to the Boolean prime ideal theorem. Conversely, in the special case of a ring $A=\mathbb{F}_2^S$ for some set $S$, compactness of $\operatorname{Spec} A$ is exactly the statement that any collection of subsets of $S$ with the finite intersection property is contained in an ultrafilter on $S$. This statement for arbitrary $S$ is equivalent to the Boolean prime ideal theorem.