A remark about the one-dimensional case. By the characterization of Krull dimension given by T. Coquand and H. Lombardi, dim($R$)$\,\leq 1$ iff $\forall_{x,y\in R}\exists_{a,b\in R,m,n\in\mathbb{N}}\,x^{m}(y^{n}(1+by)+ax)=0$. Here we may assume that neither $x$ nor $y$ is nilpotent or a unit.

If $R$ is reduced, we can require $m=1$.

If $R$ is a domain, the condition simplifies to $\exists_{b\in R,n\in\mathbb{N}}\,y^{n}(1+by)\in Rx$ for all non-zero non-units $x$ and $y$ in $R$.

If, in addition, $R$ is local with maximal ideal $\mathfrak m$, one gets dim($R$)$\,\leq 1\iff\forall_{0\neq x\in \mathfrak m}\,\mathfrak m=\sqrt {Rx}$.

A remark about the one-dimensional case. By the characterization of Krull dimension given by T. Coquand and H. Lombardi, dim($R$)$\,\leq 1$ iff $\forall_{x,y\in R}\exists_{a,b\in R,m,n\in\mathbb{N}}\,x^{m}(y^{n}(1+by)+ax)=0$. Here we may assume that neither $x$ nor $y$ is nilpotent or a unit.

If $R$ is reduced, we can require $m=1$.

If $R$ is a domain, the condition simplifies to $\exists_{b\in R,n\in\mathbb{N}}\,y^{n}(1+by)\in Rx$ for all non-zero non-units $x$ and $y$ in $R$.

If, in addition, $R$ is local with maximal ideal $\mathfrak m$, one gets dim($R$)$\,\leq 1\iff\forall_{0\neq x\in \mathfrak m}\,\mathfrak m=\sqrt {Rx}$. But this is of course quite trivial.