We may assume without loss of generality that $n_1 \ge n_2$. Since $0 < a_1 b_2 < k \cdot k^{n_2+1} = k^{n_2+2}$, we can represent
$$a_1 b_2 = \sum_{j=0}^{n_2+1} d_j k^j$$
by choosing $d_j$ with $0 \le d_j \le k-1$ ($j = 0, 1, \dots, n_2+1$) suitably.
Set $e_j = d_0 + d_1 + \dots + d_j$ ($j = 0, 1, \dots, n_2+1$).
Now let us separate the two cases:
* When $n_1 > n_2$.
We have
$$b_1 b_2 = \sum_{j=0}^{n_1} k^j \times \sum_{j=0}^{n_2+1} d_j k^j \equiv \sum_{j=0}^{n_2+1} e_j k^j \pmod{k^{n_2+2}}$$
Since $0 \le e_0 = d_0 \le k-1$, we get $a_3 = e_0$ and $a_3 \equiv e_1 \pmod k$. Consequently, we see that $d_1 = e_1 - e_0$ is a multiple of $k$, and from $0 \le d_1 < k-1$, we must have $d_1 \ne 0$. This in turn implies that $e_0 = e_1 = a_3$ and hence $e_1 = a_3 \equiv e_2 \pmod k$, and in the same way as above we conclude that $d_2 = 0$.
Repeating this argument we obtain that $d_1 = d_2 = \cdots = d_{n_2+1} = 0$, which implies that $a_1 b_2 = d_0 \le k-1$, which, in turn, contradicts the assumption that $n_1, n_2, a_1, a_2$ are all positive integers. Thus, we see that there are no triples $(n_1, n_2, n_3)$ which satisfy the requirements in this case.
* When $n_1 = n_2$.
In this case
$$b_1 b_2 \equiv \sum_{j=0}^{n_1} k^j \times \sum_{j=0}^{n_2+1} d_j k^j \equiv \sum_{j=0}^{n_2} e_j k^j \pmod{k^{n_2+1}}$$
is valid, and we can conclude that $d_1 = d_2 = \cdots = d_{n_2} = 0$ hold as in the case where $n_1 > n_2$. If we let $a_1 a_2 = c_0 + c_1 k$ ($0 \le c_0, c_1 \le k-1$), then we have
$$a_1 b_2 = c_0 + (c_0 + c_1)(k + k^2 + \cdots + k^{n_2}) + c_1 k^{n_2+1}.$$
Since $c_0 + c_1$ is a multiple of $k$ as $d_1 = 0$, and since $0 < c_0 + c_1 \le 2k-2$, we have $c_0 + c_1 = k$. If we assume that $n_1 = n_2 \ge 2$, then we get $d_2 = 1$, contradicting $d_2 = 0$. Thus, we get $n_1 = n_2 = 1$, and $n_3 = 2n_1 + 1 = 3$.
Conversely, if $(n_1, n_2, n_3) = (1, 1, 3)$, we will show that $(k, a_1, a_2, a_3)$ which satisfy the requirement exist. Since we have
$$a_1(1+k) \times a_2(1+k) = a_1 a_2(k+1)^2, \quad a_3(1+k+k^2+k^3) = a_3(k+1)(k^2+1),$$
if we can find $a_1, a_2$ satisfying $a_1 a_2 = \frac{k^2+1}{2}$, $1 \le a_i \le k-1$ ($i=1, 2$) for some odd integer $k$, then such a pair $a_1, a_2$ together with $a_3 = \frac{k+1}{2}$ will satisfy the desired conditions. By checking odd numbers in turn starting with $3$, we find that the quadruple $(k, a_1, a_2, a_3) = (7, 5, 5, 4)$ satisfies the requirement. Thus, the only triple $(n_1, n_2, n_3)$ to satisfy the requirement of the problem is $(1, 1, 3)$.

If we rewrite the relation $b_1 b_2 = b_3$ by using the summation formula for the geometric series, we get
$$a_1 a_2 \cdot \frac{k^{n_1+1}-1}{k-1} \times \frac{k^{n_2+1}-1}{k-1} = a_3 \cdot \frac{k^{n_3+1}-1}{k-1}.$$
By multiplying both sides of the equation above by $(k-1)^2$, we get
$$a_1 a_2 (k^{n_1+1} - 1) (k^{n_2+1} - 1) = a_3 (k^{n_3+1} - 1) (k-1) \quad (*)$$
Since $k^{n_2+1} - 1 \ge k^2 - 1 > a_3(k-1)$ imply that $k^{n_1+1} - 1 < k^{n_3+1} - 1$ holds, we get $n_1 < n_3$. We similarly get $n_2 < n_3$. We can also assume that $n_1 \ge n_2$ without loss of generality.
Assume that $n_2 \ge 2$. Considering the equation (*) in $\mod k^3$, we get $a_1 a_2 \equiv -a_3(k-1) \pmod{k^3}$. However, since
$$0 < a_1 a_2 + a_3 (k-1) < k^2 + k(k-1) < 2k^2 \le k^3$$
hold, we get a contradiction. Thus, we conclude that $n_2 = 1$.
If we then divide both sides of the equation (*) by $k-1$, we obtain
$$a_1 a_2 (k^{n_1+1} - 1)(k+1) = a_3 (k^{n_3+1} - 1).$$
Suppose now $n_1 \ge 2$. If we consider both sides of the equation above in $\mod k^3$, then we get $-a_1a_2(k+1) \equiv -a_3 \pmod{k^3}$. However, since
$$\begin{aligned} a_1a_2(k+1) - a_3 &\ge (k+1) - (k-1) = 2 > 0, \\ a_1a_2(k+1) - a_3 &\le (k-1)^2(k+1) - 1 = k^3 - k^2 - k < k^3 \end{aligned}$$
hold, we get a contradiction. Therefore, we must have $n_1 = 1$.
Summarizing we get
$$a_1a_2(k^2 - 1)(k + 1) = a_3(k^{n_3+1} - 1),$$
and furthermore, since
$$\begin{aligned} k^{n_3+1} - 1 &\le a_3(k^{n_3+1} - 1) = a_1a_2(k^2 - 1)(k + 1) \\ &\le (k-1)^2(k^2 - 1)(k + 1) \\ &= k^5 - k^4 - 2k^3 + 2k^2 + k - 1 \\ &< k^5 - 1, \end{aligned}$$
we obtain $n_3 < 4$.
If we suppose $n_3 = 2$, then we get $a_1a_2(k+1)^2 = a_3(k^2+k+1)$, so we conclude that $0 < a_1a_2 < a_3 < k$ holds, since $(k+1)^2 > k^2+k+1$. However, if we consider both sides of $a_1a_2(k+1)^2 = a_3(k^2+k+1)$ in $\mod k$, we get $a_1a_2 \equiv a_3 \pmod{k}$, which gives a contradiction. As $n_3 > n_1 = 1$, we conclude from $b_1b_2 = b_3$ that $(n_1, n_2, n_3) = (1, 1, 3)$. We can get $(k, a_1, a_2, a_3)$ satisfying the conditions of the problem when $(n_1, n_2, n_3) = (1, 1, 3)$ as in the preceding solution to the problem.