Suppose on the contrary that there exist 40 distinct positive integers in arithmetic progression such that each term can be written as $2^k + 3^l$, and denote this sequence by $a, a+d, a+2d, \ldots, a+39d$, where $a, d$ are positive integers. Let
$$m = \lfloor \log_2 (a + 39d) \rfloor, \quad n = \lfloor \log_3 (a + 39d) \rfloor.$$
In what follows, we first show that at most one of $a+26d, a+27d, \ldots, a+39d$ cannot be written as $2^m + 3^l$ or $2^k + 3^n$ ($k, l$ are nonnegative integers).
Suppose that $a+hd$ cannot be written as $2^m + 3^l$ or $2^k + 3^n$, for some $26 \le h \le 39$. Then, by assumption, $a+hd = 2^b + 3^c$ for some nonnegative integers $b, c$. By the definition of $m$ and $n$, it is clear that $b \le m, c \le n$. Since $a+hd$ cannot be written as $2^m + 3^l$ or $2^k + 3^n$, we have $b \le m-1, c \le n-1$.
If $b \le m-2$, then
$$\begin{aligned} a + hd &\le 2^{m-2} + 3^{n-1} = \frac{1}{4} \times 2^m + \frac{1}{3} \times 3^n \\ &\le \frac{7}{12} \times (a + 39d) < a + 26d, \end{aligned}$$
a contradiction.
If $c \le n-2$, then
$$\begin{aligned} a + hd &\le 2^{m-1} + 3^{n-2} = \frac{1}{2} \times 2^m + \frac{1}{9} \times 3^n \\ &\le \frac{11}{18} \times (a + 39d) < a + 26d, \end{aligned}$$
also a contradiction.
It follows that $b = m - 1, c = n - 1$, which implies that at most one of $a + 26d, a + 27d, \dots, a + 39d$ cannot be written as $2^m + 3^l$ or $2^k + 3^n$.
In these 14 numbers, at least 13 numbers can be written as $2^m + 3^l$ or $2^k + 3^n$. By the pigeonhole principle, at least 7 numbers can be written in the same form. We shall discuss two cases.
Case 1: There are 7 numbers in the form of $2^m + 3^l$, denoted by
$$2^m + 3^{l_1}, 2^m + 3^{l_2}, \dots, 2^m + 3^{l_7},$$
where $l_1 < l_2 < \dots < l_7$. Thus, $3^{l_1}, 3^{l_2}, \dots, 3^{l_7}$ are the 7 terms of an arithmetic progression with 14 terms and the common difference $d$. However,
$$13d \ge 3^{l_7} - 3^{l_1} \ge (3^5 - \frac{1}{3}) \times 3^{l_2} > 13(3^{l_2} - 3^{l_1}) \ge 13d,$$
a contradiction.
Case 2: There are 7 numbers in the form of $2^k + 3^n$, denoted by
$$2^{k_1} + 3^n, 2^{k_2} + 3^n, \dots, 2^{k_7} + 3^n,$$
where $k_1 < k_2 < \dots < k_7$. Thus $2^{k_1}, 2^{k_2}, \dots, 2^{k_7}$ are the 7 terms of an arithmetic progression with 14 terms and the common difference $d$. However,
$$13d \ge 2^{k_7} - 2^{k_1} \ge (2^5 - \frac{1}{2}) \times 2^{k_2} > 13(2^{k_2} - 2^{k_1}) \ge 13d,$$
a contradiction.
It follows from the above arguments that our assumption at the very beginning is false, which completes the proof.