Is there a collection of distinct positive integers $(k_1, k_2, k_3, p_1, p_2, p_3)$ such that:
$p_1, p_2, p_3$ are odd primes, and $k_1, k_2, k_3$ are odd
$(k_1 + 2) p_1 = k_2 p_2$ and $(k_2 + 2) p_2 = k_3 p_3$
$k_i \geq p_i$
Additionally, is it possible that there are infinitely many such collections? Or, that there are collections with arbitrarily many $p_i$ and $k_i$? What if we restrict to cases where $p_i < p_{i+1}$?
Note that $p_1\mid k_2$ and $p_2 \mid k_3$. Let $k_2 = a_1p_1, k_3 = a_2p_2$.
Now the requirement that $k_3 \ge p_3$ requires that $a_2 \ge p_3/p_2$. That $k_2 \ge p_2$ requires $a_1 \ge p_2/p_1$, but substituting the formulas for $a_1$ and $k_2$ gives $a_2p_3 - 2 \ge p_2$ or $a_2 \ge \frac{p_2 - 2}{p_3}$. And the requirement that $k_1 > p_1$ requires that $p_1$ be chosen so that $$p_1 \le 1 + \sqrt{1 + k_2p_2}$$ Since $1 + k_2p_2 > k_2$, we can be sure that such $p_1$ always exist.
So meeting your original conditions just requires choosing two arbitrary primes $p_2, p_3$, then choosing $$a_2 \ge \max\left\{\frac{p_3}{p_2}, \frac{p_2 -2}{p_3}\right\}$$ Then calculating $k_2$ and choosing a prime fractor $p_1$ that also satisfies $$p_1 \le 1 + \sqrt{1 + k_2p_2}$$
Since $p_2, p_3$ are arbitrary, enforcing $p_3 > p_2$ is not restrictive, and it seems obvious that choosing $p_1 < p_2$ is not difficult either.
So there are infinitely many solutions.