1. Initial Setup and Monic Polynomial Definition:
We are given that $P(x)$ is a monic polynomial such that $P(x)^2 - 1$ is divisible by $P(x+1)$. A monic polynomial is a polynomial whose leading coefficient is 1.

2. Case for Degree 0 and 1:
- For $\deg(P(x)) = 0$, $P(x)$ is a constant polynomial. Let $P(x) = c$. Then $P(x)^2 - 1 = c^2 - 1$. For $P(x+1) = c$ to divide $c^2 - 1$, $c$ must be $\pm 1$. Since $P(x)$ is monic, $c = 1$. Thus, $P(x) = 1$.
- For $\deg(P(x)) = 1$, let $P(x) = x + b$. Then $P(x+1) = x + b + 1$ and $P(x)^2 - 1 = (x + b)^2 - 1 = (x + b - 1)(x + b + 1)$. Clearly, $P(x+1) = x + b + 1$ divides $(x + b - 1)(x + b + 1)$.

3. **Claim: All Roots of $P(x)$ are the Same**:
Suppose $P(x)$ has degree $n \ge 2$. We can write $P(x) = \prod_{i=1}^n (x - z_i)$ where $z_i$ are the roots of $P$. Without loss of generality, assume $\text{Re}(z_1) \le \text{Re}(z_2) \le \dots \le \text{Re}(z_n)$.

4. **Evaluating $|P(z_i - 1)| = 1$**:
Since $P(x+1)$ divides $P(x)^2 - 1$, for any root $z_i$ of $P$, $P(z_i - 1) = 0$. Thus, $|P(z_i - 1)| = 1$. We have:
$$1 = |P(z_1 - 1)| = \left| \prod_{i=1}^n (z_1 - 1 - z_i) \right| = \prod_{i=1}^n |z_1 - 1 - z_i|$$
Since $\text{Re}(z_1 - z_i - 1) \le -1$, we have $|z_1 - z_i - 1| \ge 1$. Therefore:
$$\prod_{i=1}^n |z_1 - 1 - z_i| \ge \prod_{i=1}^n 1 = 1$$
Equality must hold, implying $|z_1 - z_i - 1| = 1$ for all $i$. This forces $z_1 = z_i$ for all $i$, meaning all roots are the same.

5. **Form of $P(x)$**:
Since all roots are the same, $P(x) = (x - c)^n$ for some constant $c$.

6. Divisibility Condition:
We need $P(x+1) = (x - c + 1)^n$ to divide $P(x)^2 - 1 = ((x - c)^n)^2 - 1$. Notice:
$$P(x)^2 - 1 = ((x - c)^n)^2 - 1 = ((x - c)^n - 1)((x - c)^n + 1)$$
Since $\gcd((x - c)^n - 1, (x - c)^n + 1) = 1$, $P(x+1)$ must divide either $(x - c)^n - 1$ or $(x - c)^n + 1$. Given the degrees, this implies:
$$P(x+1) = P(x) + 1 \quad \text{or} \quad P(x+1) = P(x) - 1$$

7. **Polynomials Satisfying $P(x+1) = P(x) \pm 1$**:
The only polynomials satisfying $P(x+1) - P(x) = 1$ or $P(x+1) - P(x) = -1$ are linear polynomials. Thus, $P(x) = x + b$ for some $b \in \mathbb{Z}$.

Conclusion:
The only monic polynomials $P(x)$ such that $P(x)^2 - 1$ is divisible by $P(x+1)$ are $P(x) = 1$ and $P(x) = x + b$ for any $b \in \mathbb{Z}$.

The final answer is $\boxed{ P(x) = 1 }$ and $P(x) = x + b$ for any $b \in \mathbb{Z}$.