$\def\bbZ{\mathbb{Z}}
\def\frm{\mathfrak{m}}
\def\bbQ{\mathbb{Q}}$(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 $A$.

For an explicit example, consider $K=\bbQ((x))$, $A=\bbQ[[x]]$ and $\mathcal{R}=\bbZ_{(p)}$, where $p\in\bbZ$ is a prime.