Is there a number field $K$ and a unit $u \in \mathcal{O}_K^{\times}$ of infinite order which can be written as an arbitrarily high power of another unit?
I think the answer is no, because a Dedekind ring is finitely generated. However, I am not sure if this is the correct line of argument.
On
Consider a number field of degree $N$. Let $0\ne u\in\mathcal O$ such that for all $m\in\mathbb N$ there exixts $v_m\in\mathcal O$ with $u=v_m^m$. Then $(u)\subseteq (v_2)\subseteq (v_4)\subseteq(v_8)\subseteq\ldots$ must be eventually stationary, so $v_{2^r}$ and $v_{2^{r+1}}$ differ by a unit, say $v_{2^r}=wv_{2^{r+1}}$, which implies that $u=v_{2^{r+1}}^{2^{r+1}}$ and $u^2=v_{2^r}^{2^{r+1}}$ differ by a unit, i.e., $u=w^{2^{r}}\in\mathcal O^\times$. (You had already found this).
Let $f(X)$ be the minimal polynomial of $u$ and assume $|u'|<R$ for all conjugates $u'$ of $u$. Then the conjugates $v'$ of $v_m$ are like $v_m$ roots of $f(X^m)$ so that $|v'|^m\le R$, i.e. $|v'|\le\sqrt[m]R$. Then for $m$ large enough all conjugates have absolute value $<2$. We conclude that the minimal polynomial of $v_m$ is of the form $$X^d+a_{d-1}X^{d-1}+\ldots +a_2X^2+a_1X\pm 1$$ with $d\le N$ and - as $a_i$ is the sum of certain products of certain subsets of the conjugates - $|a_i|\le 2^{2N}$. Since there are only finitely many such polynomials and each has only finitely many roots, the $v_m$ cannot all be distinct. But if $v_m=v_k$ with $m\ne k$ then $u^k=v_m^{mk}=v_k^{mk}=u^m$, so $u$ has finite order.
No. By Dirichlet's unit theorem the group of units is always finitely generated, and an element of infinite order in a finitely generated abelian group can never satisfy this condition.