Is there a countable field such that the multiplicative group of that field is isomorphic to $(\mathbb{Z},+)$?



Multiplicative group of countable fields
I’m going to say no. You want the non-zero elements to be $\cdots,x^{-1},1,x,x^2,\cdots$ where $x^k \neq 1$ for $k \neq 0.$ 


If the characteristic is a finite number $p$ then $1+x=x^j$ for some $j.$ If $j \gt 1$ this only allows $p^j$ elements. In case $j=-t \lt 0$ we have $x^t+x^{t+1}=1$ and $p^{t+1}$ elements.


In case of infinite characteristic every rational number is $x^j$ for a unique $j.$ In case that is not convincing enough, $2=x^j$ makes the field $\mathbb{Z}[\sqrt[j]{2}]$ which does not have the desired property, even for $j=\pm1.$


