We will prove the requirement by induction on $k$.

We notice that
$$\frac{1}{1+ma} + \frac{1}{(1+ma)(1+2ma)} = \frac{2}{1+2ma},$$
therefore there is an arithmetic progression consisting of three terms of the sequence: $x_m, x_{2m}, x_{2m^2a+3m}$.

We assume that there are $k$ non-zero natural numbers $n_1 < n_2 < \dots < n_k$ such that the numbers $x_{n_1}, x_{n_2}, \dots, x_{n_k}$ are the consecutive terms of an arithmetic progression. Considering $y = 2x_{n_1} - x_{n_2}$, the numbers $y, x_{n_1}, x_{n_2}, \dots, x_{n_k}$ are in arithmetic progression (in the number of $k+1$).

We have:
$$y = \frac{2}{1+n_1a} - \frac{1}{1+n_2a} = \frac{1+pa}{(1+n_1a)(1+n_2a)} = \frac{1+pa}{1+(n_1+n_2+n_1n_2a)a},$$
where $p = 2n_2 - n_1$.

Therefore
$$\frac{y}{1+pa}, \frac{x_{n_1}}{1+pa}, \frac{x_{n_2}}{1+pa}, \dots, \frac{x_{n_k}}{1+pa}$$
are $k+1$ terms $x_{m_1}, x_{m_2}, \dots, x_{m_{k+1}}$ of the sequence $(x_n)_{n \ge 1}$.

They are in arithmetic progression, because when we divide the terms of an arithmetic progression by a non-zero real number, we still get an arithmetic progression.

In addition, we have $m_1 < m_2 < \dots < m_{k+1}$, because the sequence $(x_n)_{n \ge 1}$ is strictly monotone. With this, the proof is complete.

We notice that $(1+na)(1+ma) = 1 + (mn + m + n)a$, for any nonzero natural numbers $m$ and $n$.

For $k \ge 3$ we consider the arithmetic progression $1+a, 1+2a, \dots, 1+ka$. We divide each term by their product, thus obtaining $k$ terms of the sequence $(x_n)_{n \ge 1}$ in arithmetic progression, because when we divide the terms of an arithmetic progression by a non-zero real number, we still get an arithmetic progression.