Here is a short and self-contained proof.
Fix an integer $k\ge 1$. It suffices to show that for each prime power $p^e$ dividing $k$ there is an index $A_{p,e}$ so that every CA number $m$ with index $>A_{p,e}$ satisfies $v_{p}(m)\ge e$. Taking $A=\max_{p^e|k}A_{p,e}$ gives the theorem.
Recall the defining property: $m$ is CA iff there exists $\varepsilon > 0$ such that
$$
F(n,\varepsilon):=\frac{\sigma(n)}{n^{1+\varepsilon}}
$$
is maximized at $n=m$. For each CA number $m$ choose one such $\varepsilon = \varepsilon_m>0$.
We want to show that $\varepsilon_m \to 0$ as $m\to\infty$. For fixed $\varepsilon_0 >0$, we have
$$
F(n,\varepsilon_0)=\frac{\sigma(n)}{n}\cdot n^{-\varepsilon_0}\le \tau(n)\,n^{-\varepsilon_0}
$$
and since $\tau(n)=n^{o(1)}$, we have
$$\tau(n)\, n^{-\varepsilon_0} \xrightarrow{n\to\infty} 0$$
and hence $F(\cdot,\varepsilon_0)$ achieves its global maximum at some finite $n$, so only finitely many CA numbers can arise as maximizers for $\varepsilon\ge\varepsilon_0$. Since this holds for every $\varepsilon_0 > 0$, the chosen $\varepsilon_m$ for CA numbers must tend to $0$ along any subsequence with $m\to\infty$.
Now, fix a prime $p$ and an exponent $e\ge 1$. Assume, toward a contradiction, that there are infinitely many CA numbers $m$ with $v_p(m) = v < e$. For such an $m$ let $\varepsilon=\varepsilon_m$ (so $\varepsilon$ can be taken arbitrarily small along the infinite family). Consider
$$
m':=m\cdot p^{e-v}
$$
so that $v_p(m') = e$. Also, we have
$$
\frac{F(m',\varepsilon)}{F(m,\varepsilon)}
= \frac{\sigma(p^{e})/p^{e(1+\varepsilon)}}{\sigma(p^{v})/p^{v(1+\varepsilon)}}
= \frac{\sigma(p^{e})}{\sigma(p^{v})}\,p^{-(e-v)(1+\varepsilon)}
$$
and hence
$$
\lim_{\varepsilon\to 0}\frac{F(m',\varepsilon)}{F(m,\varepsilon)}
= \frac{p^{e+1}-1}{p^{v+1}-1}\,p^{-(e-v)}
$$
and
$$
\frac{p^{e+1}-1}{p^{v+1}-1} - p^{e-v}
= \frac{p^{e+1}-1 - p^{e-v}(p^{v+1}-1)}{p^{v+1}-1}
= \frac{p^{e-v}-1}{p^{v+1}-1}>0
$$
implying
$$\frac{\sigma(p^{e})}{\sigma(p^{v})}\,p^{-(e-v)}>1$$
and hence, for $\varepsilon$ sufficiently small, we have $F(m',\varepsilon) >F(m,\varepsilon)$. This contradicts the fact that $m$ maximizes $F(\cdot,\varepsilon)$. Therefore only finitely many CA numbers can have $v_p<e$ and equivalently, for all sufficiently large CA indices we have $v_p\ge e$.
Putting these finite bounds together for every prime power dividing $k$ proves: there is $A$ so that for every CA number $a(n)$ with $n>A$ we have $k\mid a(n)$. This completes the proof.
