How many nonzero coefficients can a polynomial $ P(x)$ have if its coefficients are integers and $ |P(z)| \le 2$ for any complex number $ z$ of unit length?
1. **Parseval's Theorem Application**:
   We start by applying Parseval's theorem, which states that for a polynomial \( P(x) \) with complex coefficients, the sum of the squares of the absolute values of its coefficients is equal to the integral of the square of the absolute value of the polynomial over the unit circle. Mathematically, this is expressed as:
   \[
   \int_{0}^{2\pi} |P(e^{it})|^2 \, dt = 2\pi \sum_{n=-\infty}^{\infty} |a_n|^2
   \]
   Given that \( |P(z)| \leq 2 \) for any complex number \( z \) of unit length, we have:
   \[
   \int_{0}^{2\pi} |P(e^{it})|^2 \, dt \leq \int_{0}^{2\pi} 2^2 \, dt = 8\pi
   \]
   Therefore:
   \[
   2\pi \sum_{n=-\infty}^{\infty} |a_n|^2 \leq 8\pi \implies \sum_{n=-\infty}^{\infty} |a_n|^2 \leq 4
   \]

2. **Nonzero Coefficients Analysis**:
   Since the coefficients \( a_n \) are integers, the sum of their squares being at most 4 implies that there can be at most 4 nonzero coefficients. If there were exactly 4 nonzero coefficients, each would have to be \( \pm 1 \) to satisfy the sum of squares condition:
   \[
   |a_1|^2 + |a_2|^2 + |a_3|^2 + |a_4|^2 = 4
   \]
   This implies:
   \[
   |a_i| = 1 \text{ for } i = 1, 2, 3, 4
   \]

3. **Contradiction for 4 Nonzero Coefficients**:
   If \( P(z) \) had exactly 4 nonzero coefficients, then by Parseval's theorem:
   \[
   \int_{0}^{2\pi} |P(e^{it})|^2 \, dt = 8\pi
   \]
   Given the hypothesis \( |P(z)| \leq 2 \) for \( |z| = 1 \), this would imply \( |P(z)| = 2 \) for all \( |z| = 1 \). This leads to:
   \[
   P(z)P(1/z) = 4 \text{ for } |z| = 1
   \]
   Extending this to all \( z \), we get:
   \[
   P(z)P(1/z) = 4
   \]
   This implies that the only zero of \( P(z) \) is at \( z = 0 \), leading to \( P(z) = a z^n \) for some integer \( a \) and \( n \). However, this contradicts the assumption of having 4 nonzero coefficients.

4. **Analysis for 3 Nonzero Coefficients**:
   Suppose \( P(z) \) has exactly 3 nonzero coefficients. We can write:
   \[
   P(z) = 1 + az^k + bz^l \quad \text{with} \quad k < l \quad \text{and} \quad a, b \in \{1, -1\}
   \]
   Let \( z_j \) be the roots of \( bz^l = -1 \). Then:
   \[
   |2 + az_j^k| \leq 2
   \]
   Summing over all roots \( z_j \), we get:
   \[
   \sum_{j} |2 + az_j^k| = 2l
   \]
   This implies \( |2 + az_j^k| = 2 \) for all \( j \), leading to \( 2 + az_j^k = 2 \), which is impossible.

5. **Conclusion**:
   Therefore, \( P(z) \) can have at most 2 nonzero coefficients. This is possible, for example, with \( P(z) = 1 + z \) or \( P(z) = 1 - z \).

The final answer is \(\boxed{2}\)