Find all positive integers $d$ with the following property:  there exists a polynomial $P$ of degree $d$ with integer coefficients such that $\left|P(m)\right|=1$ for at least $d+1$ different integers $m$.
1. **Identify the problem constraints:**
   We need to find all positive integers \( d \) such that there exists a polynomial \( P \) of degree \( d \) with integer coefficients, and \(|P(m)| = 1\) for at least \( d+1 \) different integers \( m \).

2. **Construct examples for small \( d \):**
   - For \( d = 1 \), consider \( P(x) = 2x - 1 \). This polynomial satisfies \(|P(1)| = 1\) and \(|P(0)| = 1\), so it works.
   - For \( d = 2 \), consider \( P(x) = 2x(x-2) + 1 \). This polynomial satisfies \(|P(0)| = 1\), \(|P(1)| = 1\), and \(|P(2)| = 1\), so it works.
   - For \( d = 3 \), consider \( P(x) = x(x-1)(x-3) + 1 \). This polynomial satisfies \(|P(0)| = 1\), \(|P(1)| = 1\), \(|P(2)| = 1\), and \(|P(3)| = 1\), so it works.

3. **Prove that \( d \geq 4 \) fails:**
   - We need at least \( d+1 \) total integer roots of \( P(x) - 1 \) and \( P(x) + 1 \). Without loss of generality, assume \( P(x) + 1 \) has at least \(\frac{d+1}{2}\) of these roots.
   - By shifting vertically, the problem is equivalent to having at least \( d+1 \) total integer roots of \( P(x) \) and \( P(x) - 2 \), with at least \(\frac{d+1}{2}\) total integer roots of \( P(x) \).

4. **Analyze the case \( d \geq 4 \):**
   - Since \( d \geq 4 \), \( P(x) \) must have at least 3 integer roots. Write \( P(x) = (x-a)(x-b)(x-c)Q(x) \) for some nonzero polynomial \( Q \) and distinct integers \( a, b, c \).
   - If \( P(x) = 2 \), then we require \(\{x-a, x-b, x-c\} = \{2, 1, -1\}\) or \(\{1, -1, -2\}\). Since \( a, b, c \) are distinct, we cannot have \( x-a = x-b = 1 \) and \( x-c = 2 \).

5. **Consider the case \( P(x) - 2 \) has at least 2 integer roots:**
   - If \( P(x) - 2 \) has at least 2 integer roots, then \( P(x) \) still has at least 3 integer roots. Write \( P(x) = (x-a)(x-b)(x-c)Q(x) \).
   - For \( P(x) = 2 \), we need \(\{x-a, x-b, x-c\} = \{2, 1, -1\}\) or \(\{1, -1, -2\}\). This leads to contradictions as shown in the solution.

6. **Consider the case \( P(x) - 2 \) has exactly 1 integer root:**
   - If \( P(x) - 2 \) has exactly 1 integer root, then \( P(x) \) must have at least 4 integer roots. Write \( P(x) = (x-a)(x-b)(x-c)(x-d)Q(x) \).
   - For \( x \) to be a root of \( P(x) - 2 \), we need \( x-a, x-b, x-c, x-d \in \{1, -1, 2, -2\} \). By the Pigeonhole Principle, \(\{x-a, x-b, x-c, x-d\} = \{1, -1, 2, -2\}\), but then \( 4 \mid P(x) \), leading to a contradiction.

7. **Consider the case \( P(x) - 2 \) has no integer roots:**
   - If \( P(x) - 2 \) has no integer roots, then all \( d+1 \) roots must belong to \( P(x) \). However, \( P(x) \) has degree exactly \( d \), so this is absurd. The only polynomial of degree at most \( d \) which has at least \( d+1 \) roots is the zero polynomial, which cannot have degree \( d \).

8. **Conclusion:**
   - Thus, \( d \geq 4 \) fails, and the only possible values for \( d \) are \( d \leq 3 \).

The final answer is \( \boxed{ d \leq 3 } \).