In general, if $R$ is a ring of characteristic $p > 0$, write $F_*^n R$ for the ring $R$ but viewed as $R$-algebra via the $p^n$-power Frobenius $R \to F_*^nR$, $x \mapsto x^{p^n}$. Define the perfection $R^{\text{perf}}$ as the colimit of the tower
$$R \to F_*R \to F_*^2R \to \ldots.$$
It has a natural map from $R$ inserting into the $0^{\text{th}}$ term. For instance, if $k$ is a field of characteristic $p$ and $R = k[x_1,\ldots,x_n]$, then $R^{\text{perf}} = k^{\text{perf}}\bigl[x_1^{1/p^\infty},\ldots,x_n^{1/p^\infty}\bigr]$.
Lemma. The map $\phi \colon \operatorname{Spec} R^{\text{perf}} \to \operatorname{Spec} R$ is a homeomorphism.
In particular, $\operatorname{Spec} R^{\text{perf}}$ is topologically Noetherian if and only if $\operatorname{Spec} R$ is.
Proof. For any perfect field $k$ (in particular if $k$ is algebraically closed), the map $\operatorname{Hom}(R^{\text{perf}},k) \to \operatorname{Hom}(R,k)$ is a bijection, by the universal property of perfection (or: check that you can extend $f \colon R \to k$ uniquely to $R^{\text{perf}}$ by sending $r \in F_*^nR$ to $\phi(r)^{1/p^n}$). This implies that $\phi$ is a bijection. Since $R \to R^{\text{perf}}$ is integral, the map $\phi$ is closed [Tags 00HZ and 00GU], so it is a homeomorphism. $\square$
(In fact, with a little bit more work one shows that such maps are universal homeomorphisms: it remains a homeomorphism after arbitrary base change. See [Tags 04DF and 01S2] for details.)
