China Western Mathematical Olympiad

For any given positive integer $k$, choose positive integer $m$ sufficiently large such that $2^m + 3^m - k > 1$. Consider the following $k$ integers:
$$2^m + 3^m - 1, 2^m + 3^m - 2, \dots, 2^m + 3^m - k,$$
all of which are larger than $1$. From each of these integers, pick a prime factor: $p_1, p_2, \dots, p_k$, and let
$$n_t = m + t(p_1 - 1)(p_2 - 1)\cdots(p_k - 1),$$
where $t$ is an arbitrary positive integer. For any fixed integer $i$ ($1 \le i \le k$), one has $2^{n_t} \equiv 2^m \pmod{p_i}$. In fact, if $p_i = 2$, then the result is obvious. Assuming that $p_i \ne 2$, it follows from Fermat's little theorem that
$$2^{n_t} = 2^m \cdot 2^{t(p_1-1)(p_2-1)\cdots(p_k-1)} \equiv 2^m \cdot 1 = 2^m \pmod{p_i}.$$
Similarly, one has $3^{n_t} \equiv 3^m \pmod{p_i}$. Observe that
$$2^{n_t} + 3^{n_t} - i \equiv 2^m + 3^m - i \equiv 0 \pmod{p_i},$$
$$2^{n_t} + 3^{n_t} - i > 2^m + 3^m - i.$$
Hence, $2^{n_t} + 3^{n_t} - i$ is a composite number.
Therefore, $n_t$ is one of the positive integers $n$ such that
$$2^n + 3^n - 1, 2^n + 3^n - 2, \dots, 2^n + 3^n - k$$
are all composite. As $t$ is arbitrarily chosen, there are infinitely many such positive integers satisfying the conditions above.