The answer is yes. Suppose that $Q(x, y, z)$ is a polynomial with integer coefficients such that for all integers $x, y, z$, we have

* $Q(x, y, z) \ge 0$,
* if $Q(x, y, z) = 0$ then $x$ is a non-square, and
* for each positive non-square $x$, there exist $y, z \in \mathbb{Z}$ with $Q(x, y, z) = 0$.

We claim that the polynomial
$$P(x, y, z) = x - xQ(x, y, z)$$
satisfies the desired property. Indeed, $P(x, y, z) \le 0$ unless $Q(x, y, z) = 0$, in which case $x$ must be a non-square, so if $P(x, y, z)$ is a positive integer, then it is a non-square. Further, for any positive non-square $x$, choosing $y, z$ for which $Q(x, y, z) = 0$ yields a triple $(x, y, z)$ for which $P(x, y, z) = x$. It remains only to show that such a polynomial $Q(x, y, z)$ exists. We present two approaches.

First approach:

For any $x \in \mathbb{Z}$, we have the following chain of five equivalent statements:
(i) $x$ is a positive non-square,
(ii) $y^2 < x < (y+1)^2$ for some $y \in \mathbb{Z}$,
(iii) $(x - y^2)((y + 1)^2 - x) > 0$ for some $y \in \mathbb{Z}$,
(iv) $((x - y^2)((y + 1)^2 - x) - z)^2 = 0$ for some $y, z \in \mathbb{Z}$,
(v) $Q(x, y, z) = 0$ for some $y, z \in \mathbb{Z}$, where
$$Q(x, y, z) = ((x - y^2)((y + 1)^2 - x) - z)^2$$
satisfies the desired conditions.

Second approach:

It is well known that the Pell equation $y^2 - xz^2 = 1$ with parameter $x$ has a positive integer solution $y, z \in \mathbb{Z}$ if and only if $x$ is not a square. Therefore the polynomial
$$Q(x, y, z) = (y^2 - xz^2 - 1)^2$$
also satisfies the desired conditions.

Remark. For any $k$, the problem statement also works with “not a $k$th power” instead of “not a square.” However, in this case the second approach breaks down because we can no longer use the theory of Pell equations.