Can we characterize all infinite PID s whose group of units is singleton?

Question

I am looking for a way to characterize all infinite PID s having exactly one unit i.e. invertible element ( finite PID s are not interesting , they are all fields ) . The only such example I know of is $\mathbb Z_2[x]$ . Towards characterizing such domains, first of all if $D$ is such a PID then it cannot be a field and its characteristic is $2$ as $1=-1$ in it, so it contains a copy of $\mathbb Z_2$. Moreover, every element of $D\setminus \mathbb Z_2$ is transcendental over $\mathbb Z_2$, since other-wise if $a \in D\setminus \mathbb Z_2$ is algebraic i.e. integral over the field $\mathbb Z_2$ ( https://en.wikipedia.org/wiki/Integral_element ) then $\mathbb Z_2[a]$ would be an inetgral domain which is an integral extension over a field ( namely $\mathbb Z_2$ ) implying that $\mathbb Z_2[a]$ is a field, but in that case $\mathbb Z_2[a]$ contains at least two units namely $1,a$, then so does $D$, contradiction ! But other than these I have not been able to derive any further consequences in the way of characterizing , and it is to be noted that in all the previous arguments we used nothing more than that $D$ is an integral domain, we haven't seriously used the PID nature of $D$ which if used is bound to give some further restrictions, but I can't figure out any. Any reference or link relating to this question will be highly appreciated. Please help. Thanks in advance. (Note that if we instead wanted such a UFD there would be plenty of examples to go about, namely polynomials in any $n$-variables over $\mathbb Z_2$ ... but seeking PID strikes out such examples.)