$\def\bbZ{\mathbb{Z}} \def\frm{\mathfrak{m}} \def\bbQ{\mathbb{Q}}$(For me, a DVR is never a field.)
As a complement to Pete L. Clark's answer, here's a recipe to construct a valuation ring with $\bbZ\times\bbZ$ as value group (the following is taken from Matsumura's Commutative Ring Theory, Remark after Theorem 11.1): Let $K$ be a field, and suppose that $A$ is a DVR of $K$. Suppose $\mathcal{R}$ is a DVR of $k=A/\frm_A$, and let $R$ be the inverse image of $\mathcal{R}$ along $\varphi:A\to k$, which is a valuation ring of $K$. Let $v_A$ (resp., $v_\mathcal{R}$) be a valuation on $K$ (resp., on $k$) with ring $A$ (resp., $R$). Let $f$ be a uniformizer for $A$. Then \begin{align*} v:K^*&\to\bbZ\times\bbZ\\ x&\mapsto(n,m),\quad n=v_A(x),\;m=v_\mathcal{R}(\varphi(xf^{-n})), \end{align*} is an onto valuation on $K$ with ring $A$. (Hint: in Matsumura's book it is argued that if $\overline{g}\in \mathcal{R}$ is a uniformizer, then $g$ is a uniformizer for $R$ and $g\in A\setminus\frm_A$. It follows that $v$ is onto since $x=f^ng^m$ satisfies $v(x)=(n,m)$, and leveraging $f$ and $g$—plus writing elements of the quotient field of a DVR as a product of a unit of the DVR times a power of some uniformizer—it is an exercise to show that $v$ is a valuation.)
For an explicit example, consider $K=\bbQ((x))$, $A=\bbQ[[x]]$ and $\mathcal{R}=\bbZ_{(p)}$, where $p\in\bbZ$ is a prime.