Is $Th(\mathbb{Z}[x])$ uncountably categorical? Consider $T=Th(\mathbb{Z}[x])$ in the language $L = \{0,1,+,\times,deg(), \circ\}$ where $0,1,+$ and $\times$ have their usual interpretations, $deg()$ is a unary function symbol which gives the degree of a polynomial and $\circ$ is a binary  function symbols where (if $p(x)$ and $q(x)$ are polynomials) $$ p(x)\circ q(x) = p(q(x))$$ and if $p(x)$ is a "constant", then $$p(x) \circ q(x) = p(x)$$
Clearly, $T$ is not countably categorical since it has $\aleph_0$ many $1-$types definable without parameters. However, I cannot figure out whether the theory is uncountably categorical. 
Thank you. 

This is a response to Rene Schipperus's comment asking whether or not we can define an ordering on the integers in this structure. It turns out that we can. 
We say that $P(x)$, i.e. "x is positive", holds on an element of our structure if $(\exists y)(deg(y)=x)$.
Note that $x$ is an "integer" if and only if $deg(x) = 0$. 
Now, we say that $x < y \iff (deg(x)=deg(y)=0)\wedge (\exists z)[(P(z)\wedge (z\neq 0) \wedge  (x+z =y)]$
