Let $R=\mathbb{k}[x,y]$ be a polynomial ring in two variables $x,y$, and consider the ideal $I = \langle x^2+y^2\rangle\subset R$.
$G\equiv \{x^2+y^2\}$ is is a Groebner basis for $I$ for any ordering, because it vacuously satisfies the S-pair criterion for the Buchberger algorithm (i.e. $I$ is a principal ideal).
Since $G$ is a Groebner basis for $I$ w.r.t., say, graded lexicographic order, the initial ideal is generated by the initial monomial of the generator of the ideal, namely $x^2$. So:
$\text{in}_\prec(I) = \langle x^2\rangle.$
In particular, the set of standard monomials (those monomials not in the initial ideal) are
$\text{std}_\prec (I) = \{1,y,y^2, y^3,\cdots , x, xy,xy^2,xy^3,\cdots\}$
In particular, the number of standard monomials is infinite. The Krull dimension of the corresponding affine variety is then read off as the degree of the Hilbert polynomial, which is equal to the number of standard monomials of degree $k$ or less (for sufficiently large $k$):
$p(k) \underset{k\to \infty}{\sim} 2k-1$
Since the Hilbert polynomial is of degree one, the Krull dimension of the ideal is equal to one:
$\dim_\text{Krull}(I) = 1.$
However, the dimension of the corresponding affine variety is zero, since $V(I) = \{(0,0)\}$ (the quadratic form is positive-definite):
$\dim_\mathbb{k}V(I)=0.$
What's going on here?
It's not true that $V(I) = \{(0, 0)\}$ if $\mathbb{k}$ is algebraically closed. Indeed, if there exists an element $i \in \mathbb{k}$ with $i^2 = -1$, then we have $$V(I) = \{(t, it): t \in \mathbb{k}\} \cup \{(t, -it): t \in \mathbb{k}\} = V(\langle x + iy\rangle) \cup V(\langle x - iy\rangle).$$ (Relatedly, $x^2 + y^2 = (x + iy)(x - iy).$) In this case, $V(I)$ is a union of lines, and one-dimensional at most of its points.