This is false. Let $K=bar{mathbb{Q}}$ and $L=bar{mathbb{F}}_2$. These are clearly both algebraically closed, of different characteristics, so $Knotcong L$. However, if we ONLY look at the topology, $mathrm{Spec}(K[x])$ and $mathrm{Spec}(L[x])$ will be be countable sets with the finite complement topology on the closed points, with a single generic point, so they're homeomorphic. For algebraically closed fields, the TOPOLOGY on the affine line over the field is determined by the cardinality.
For higher dimensions, it's less clear to me, because you might be able to recover characteristic (I'm a char 0 kind of person, so I don't know) from how the various curves/hypersurfaces sit inside it.
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