Given:

1. An infinite arithmetic sequence of integers $a_1, a_2, \ldots$ with the common difference $d = a_2 - a_1$.
2. An infinite geometric sequence of integers $g_1, g_2, \ldots$ with the common ratio $r \neq 1$.
3. The condition that $a_n - g_n$ is divisible by $m$ for all $n \geq 1$.
4. The further condition that $d$ (the common difference) is not divisible by $m$.

We need to determine for which positive integers $m$ these conditions hold.

### Analysis

Arithmetic and Geometric Sequences:
- For the arithmetic sequence, we have:
$$a_n = a_1 + (n-1)d.$$
- For the geometric sequence, we have:
$$g_n = g_1 \cdot r^{n-1}.$$

Difference Condition:
The condition $a_n - g_n \equiv 0 \pmod{m}$ implies:
$$a_1 + (n-1)d - g_1r^{n-1} \equiv 0 \pmod{m}.$$

Rewriting, this gives:
$$a_1 - g_1 \equiv (g_1r^{n-1} - (n-1)d - a_1) \equiv 0 \pmod{m}.$$

### Divisibility Argument

The first condition $a_n - g_n \equiv 0 \pmod{m}$ holds for all $n$ if and only if $a_1 \equiv g_1 \pmod{m}$ and

$$\forall n, \quad (n-1)d \equiv g_1 (r^{n-1} - 1) \pmod{m}.$$

For the second sequence property $a_2 - a_1 \not\equiv 0 \pmod{m}$, this indicates $d \not\equiv 0 \pmod{m}$.

Condition Conclusions:

- If $m$ is squarefree, then every divisor of $m$ is prime, such that $r^{n-1} - 1 \equiv 0 \pmod{p}$ (for any prime divisor $p$) forces $r \equiv 1 \pmod{p}$, contradicting the non-zero difference condition since otherwise, all differences $(n-1)d$ would need to be zero modulo $m$.

- If $m$ is not squarefree, i.e., $m$ contains a repeated prime factor $p^2$, we can choose $r \equiv 1 \pmod{p}$ but $r \not\equiv 1 \pmod{p^2}$, allowing for the divisibility of differences to meet all conditions without $d \equiv 0 \pmod{m}$.

### Conclusion

Thus, $m$ must be not squarefree for these sequences with the conditions given to exist.

$$\boxed{\text{m is not squarefree}}$$