1. Understanding the problem: We need to find the largest integer $r$ such that for any polynomial $P(x,y,z)$ with integer coefficients satisfying $P(a,b,c) = 0 \Leftrightarrow a = b = c$, the expression $m^r$ divides $P(n, n+m, n+2m)$ for all integers $m$ and $n$.

2. Analyzing a specific polynomial: Consider the polynomial $P(x,y,z) = (x-y)^2 + (y-z)^2$. This polynomial satisfies $P(a,b,c) = 0 \Leftrightarrow a = b = c$.

3. Evaluating the polynomial at specific points: Evaluate $P(x,y,z)$ at $(x,y,z) = (n, n+m, n+2m)$:
$$P(n, n+m, n+2m) = (n - (n+m))^2 + ((n+m) - (n+2m))^2 = (-m)^2 + (-m)^2 = m^2 + m^2 = 2m^2$$
This shows that $2m^2$ is always divisible by $m^2$, implying $r \leq 2$.

4. Generalizing the result: To show that $m^2$ always divides $P(n, n+m, n+2m)$ for any polynomial $P(x,y,z)$ satisfying the given condition, consider the polynomial $Q(t) = P(n, n+t, n+2t)$ for any fixed $n$.

5. **Properties of $Q(t)$**: The polynomial $Q(t)$ has exactly one root at $t = 0$ because $P(n, n, n) = 0$. Since $P(a,b,c) = 0 \Leftrightarrow a = b = c$, $t = 0$ is the only root of $Q(t)$.

6. Multiplicity of the root: Since $Q(t)$ is a polynomial with integer coefficients and $t = 0$ is its only root, the root must have a multiplicity of at least 2. This is because if the root had a multiplicity of 1, $Q(t)$ would change sign around $t = 0$, contradicting the condition that $P(a,b,c) = 0 \Leftrightarrow a = b = c$.

7. **Divisibility by $t^2$**: Therefore, $t^2$ divides $Q(t)$, implying $m^2$ divides $Q(m) = P(n, n+m, n+2m)$.

8. Conclusion: Since $m^2$ always divides $P(n, n+m, n+2m)$ for any polynomial $P(x,y,z)$ satisfying the given condition, the largest integer $r$ is 2.

The final answer is $\boxed{ r = 2 }$.