We are tasked with finding all polynomials $P(x)$ with integer coefficients such that $P(0) \neq 0$ and for all nonnegative integers $n, m$, the expression $P^n(m) \cdot P^m(n)$ is a square of an integer. The polynomial $P^n(m)$ denotes the polynomial $P$ applied iteratively $n$ times to $m$.

To solve the problem, we analyze the condition that for all nonnegative integers $n$ and $m$, $P^n(m) \cdot P^m(n)$ is a perfect square.

### Key Observations
1. Iterative Property: If $P(x) = x + 1$, then $P^n(m) = m + n$ because each iteration increases the value by 1. Thus, $P^n(m) \cdot P^m(n) = (m+n) \cdot (n+m) = (n+m)^2$, which is clearly a perfect square.

2. General Polynomial Form: Consider $P(x) = a_k x^k + a_{k-1} x^{k-1} + \cdots + a_0$, where the $a_i$ are integer coefficients.

- Degree: If $P(x)$ has a degree greater than 1, i.e., $k > 1$, then applying $P$ iteratively results in exponential growth, which makes $P^n(m) \cdot P^m(n)$ unlikely to be a perfect square for arbitrary $n, m$.

- Degree Zero Case: If $P(x)$ is constant, then $P^n(m)$ and $P^m(n)$ will be constants, leading to $P^n(m) \cdot P^m(n)$ not forming a consistent perfect square, violating the hypotheses.

### Concluding Based on Observations

Thus, the most simple and suitable polynomial of degree 1, which fits the condition for all nonnegative integers $n$ and $m$, is $P(x) = x + 1$.

Therefore, the only polynomial $P(x)$ satisfying all given conditions is:
$$\boxed{x + 1}$$
This choice works as verified above, illustrating that $P(x) = x + 1$ consistently satisfies the condition of making $P^n(m) \cdot P^m(n)$ a perfect square for all nonnegative integer values of $n$ and $m$.