Let $p_{1}, p_{2}, \cdots$ denote all prime numbers in ascending order.
Now, we construct the sequence $\left\{a_{n}\right\}$ that meets the requirements.
Let $a_{1}=2$, and suppose $a_{1}, a_{2}, \cdots, a_{n}$ have been determined. Take $a_{n+1}$ as the smallest positive integer greater than $a_{n}$ that satisfies the following system of congruences:
$$\left\{\begin{array}{l} x \equiv 0\left(\bmod p_{1}\right), \\ x \equiv-1\left(\bmod p_{2}\right), \\ \cdots \\ x \equiv-n\left(\bmod p_{n+1}\right) \end{array}\right.$$

Note that the Chinese Remainder Theorem guarantees the existence of such an $a_{n+1}$.
We claim that the sequence $\left\{a_{n}\right\}$ defined by the above recursion satisfies the problem's requirements. For any $k \in \mathbf{N}^{*}$, when $n \geqslant k+1$, we have $k+a_{n} \equiv 0\left(\bmod p_{k+1}\right)$. Combining this with the fact that $\left\{a_{n}\right\}$ is increasing, it follows that every term of $\left\{k+a_{n}\right\}$ from the $(k+2)$-th term onward is a multiple of $p_{k+1}$ and is greater than $p_{k+1}$. Therefore, $\left\{k+a_{n}\right\}$ has at most $k+1$ prime terms.

The proposition is thus proved.