Solution:

The values sought are all and only the odd positive integers. Indeed, if $h$ is even, then $a_{1}=1+h$ is odd, hence $a_{2}=1+2 h$ is odd, and so on, so all the numbers $a_{n}=1+n h$ are odd and greater than 1.
If instead $h$ is odd, we observe that all the terms of the sequence are either positive integers less than or equal to $h$ or even positive integers greater than $h$ and less than or equal to $2 h$. A simple proof of this fact can be obtained using the principle of induction. As the base case, we have $a_{0}=1$, which is a positive integer less than or equal to $h$. Assuming the claim is true for $a_{n}$, let us prove it for $a_{n+1}$: if $a_{n}$ is odd and less than or equal to $h$, then $a_{n+1}=a_{n}+h$ is even, greater than $h$ and less than or equal to $2 h$; if instead $a_{n}$ is even and $a_{n} \leq 2 h$, then $a_{n+1}=\frac{a_{n}}{2} \leq h$.
It follows that the sequence of numbers $a_{n}$ can take only finitely many values. Since the sequence is infinite, there will exist two indices $r<s$ such that $a_{r}=a_{s}$. Suppose we choose two indices $r<s$ such that $a_{r}=a_{s}$ and where $r$ is as small as possible. If $r=0$, then $a_{s}=a_{0}=1$ and the claim is proved. Let us show by contradiction that we cannot have $r>0$, distinguishing two cases:
- $a_{r} \leq h$; then it cannot be that $a_{r}=a_{r-1}+h$ or $a_{s}=a_{s-1}+h$, because in that case $a_{r-1}$ or $a_{s-1}$ would not be positive, contradicting what was stated previously. Therefore $a_{r-1}=a_{s-1}=2 a_{r}$, and thus we have an equality between two terms of the sequence in which the first term has an index smaller than $r$, a contradiction;
- $a_{r}$ even, $h<a_{r} \leq 2 h$; then it cannot be that $a_{r}=\frac{a_{r-1}}{2}$ or $a_{s}=\frac{a_{s-1}}{2}$, because otherwise $a_{r-1}>2 h$ or $a_{s-1}>2 h$, contradicting what was stated previously. Therefore $a_{r-1}=a_{s-1}=a_{r}-h$, and again we obtain the contradiction of an equality between two terms of the sequence in which the first term has an index smaller than $r$.