I claim that, for a field $K$, the following are equivalent:
(i) $K$ can be given a nontrvial norm -- i.e., there exists $x \in K$ with $|x| \neq 0,1$.
(ii) $K$ admits a nontrivial rank one valuation $v$.
(iii) $K$ admits infinitely many inequivalent rank one valuations $v$ such that $(K,v)$ is not complete.
(iv) $K$ is not an algebraic extension of a finite field.
Some of these facts are proved in
http://math.uga.edu/~pete/8410Chapter2v2.pdf
(see e.g. Theorem 1). Here is a sketch of (iv) $\implies$ (iii):
Suppose $K$ has characteristic $0$. Then $K$ contains $\mathbb{Q}$, which admits the $p$-adic valuations $v_p$. By Theorem 1 of loc. cit., each $v_p$ extends to a valuation on $K$. (If $K$ has characteristic $p$ and is not algebraic over $\mathbb{F}_p$, then $K$ contains $\mathbb{F}_p(t)$ which also carries infinitely many valuations $v_P$ corresponding to the irreducible polynomials $P \in \mathbb{F}_p[t]$.)
(F.K. Schmidt) If any field $K$ is complete with respect to two inequivalent rank one valuations, it is algebraically closed and uncountable. See e.g. Theorem 24 of
http://math.uga.edu/~pete/8410Chapter3.pdf
- So we are reduced to the case in which $K$ is algebraically closed and uncountable. Then $K$ is isomorphic to the algebraic closure of $K(t)$. If we give $K$ the trivial valuation and $K(t)$ the Gauss norm $v$, then the algebraic closure of $K(t)$ has infinite degree over $K(t)$ so any extension of $v$ to the algebraic closure is not complete. The image of the Gauss norm $v$ under the group $PGL_2(K)$ of linear fractional transformations gives us infinitely more pairwise inequivalent valuations.