Let $p$ be a prime, and let $f \in \mathbb{Z}_p[[x_1,\dots,x_d]]$ be a power series convergent on all of $\mathbb{Z}_p^d$. We make the following definition concerning the approximation of $f$ by polynomials:
Definition: Let $U \subset \mathbb{Z}_p^d$ be an open subset. We say that a polynomial $g \in \mathbb{Z}_p[x_1,\dots,z_d]$ is an adequate approximation of $f$ on $U$ if for every $x = (x_1,\dots,x_d) \in U$ there exists a unit $u_x \in 1 + p\mathbb{Z}_p$ such that $f(x) = u_x\times g(x)$. We say that $f$ is adequately approximable if there exists a covering of $\mathbb{Z}_p^d$ by open sets such that $f$ admits an adequate approximation on each.
Question: Let $f$ be as above. Is $f$ adequately approximable?
Progress: I believe that the answer is yes in the single-variable case (i.e., the case where $d = 1$). Indeed, the Weierstrass preparation theorem implies the existence of a polynomial $g(x_1) \in \mathbb{Z}_p[x_1]$ such that $f(x_1)/g(x_1) \in 1 + p\mathbb{Z}_p[[x_1]]$. In other words, in the single-variable case, $f$ admits a single adequate approximation on all of $\mathbb{Z}_p$.
As for the multivariable case, I suspect that the answer is no but am unsure of how to go about finding a counterexample. For instance, take $d = 2$. If $f$ admits a single adequate approximation on all of $\mathbb{Z}_p^2$ by a polynomial $g \in \mathbb{Z}_p[x_1,x_2]$, then $f$ and $g$ share the same vanishing locus, which means that the analytic curve cut out by $f$ is algebraic. I imagine that this is not true of all analytic curves, but the possibility remains that $f$ admits various local adequate approximations that together render $f$ adequately approximable.
I would also note that if $f$ is adequately approximable, then by compactness of $\mathbb{Z}_p^d$, we can approximate $f$ with just finitely many polynomials $g$ over a finite union of open sets.
