The answer is $1$ for $m = 0$, $p$ for $m = 1$ and $\varphi(p^m)$ for $m > 1$.

Note that $\mathbb{Z}_p[x]$ is a unique factorization domain. So any monic $P(x) \in \mathbb{Z}_p[x]$ can be uniquely expressed as $P(x) \equiv A(x)^2 Q(x) \pmod{p}$, where $A(x), Q(x)$ are monic polynomials and $Q(x)$ is square-free. Let $f_n$ be the number of all square-free monic polynomials of $\mathbb{Z}_p[x]$ with degree $n$. Any non-square-free monic polynomial can be expressed as $P(x) = A(x)^2 S(x)$ where $S(x)$ is a square-free monic polynomial.

If $\deg A(x) = m$, then $S(x) \in f_{n-2m}$. We have $f_{n-2m}$ options for $S(x)$ and $p^m$ options for $A(x)$. So
$$f_n = p^n - \sum_{m=1}^{\lfloor \frac{n}{2} \rfloor} f_{n-2m} p^m.$$
And by simple induction on $n$ we get $f_n = \varphi(p^n)$.