The asymptotic behavior of a recurrence related to stable matchings
Here is a proof. Parts of the proof involve some real analysis; I've sketched the details in an appendix, and if you know real analysis, you should be able to fill in the details fairly easily.

First, let's notice that for $b_n=a_{2^n}$, we have the recurrence

$$b_n = 3b_{n-1}^2 - 2b_{n-2}^4.$$

Now, let's assume that $b_n = r s^{2^n}$. The equation becomes

$$r s^{2^n} = 3 r^2 s^{2^n} - 2 r^4 s^{2^n},$$

and $s^{2^n}$ factors out. Solving the equation $r = 3r^2 - 2r^4$, we find two positive roots,  $r = 0.366 = \frac{1}{2}(\sqrt{3}-1)$ and $r= 1$. Let $r_0$ be the smaller of these roots.

Suppose we could show that $\frac{b_{n-1}^2}{b_n} = r_0$. Then we can show by induction that $b_n=r_0 s^{2^n}$ for some value of $s$. Suppose this is true for some value $n$. Then
\begin{align}
b_{n+1} &amp;= \frac{b_{n}^2}{r_0} = \frac{\left(r_0 s^{2^n}\right)^2}{r_0} = r_0 s^{2^{n+1}}.
\end{align}

In fact, we won't quite show this; what we will show is that $\frac{b_{n-1}^2}{b_n}$ goes to $r_0$ as $n$ goes to $\infty$. With some analysis (given in an appendix), we can complete the proof that $b \approx r_0 s^{2^n}$ for some value of $s$. The actual value of $s$ depends on the initial conditions, and the only way I know to get it is computationally. Maple seems to say that $s$ is around 2.280142. If you start with $a_1=2$, $a_2=7$, Maple gives $s \approx $ 4.189425.

Why does $\frac{b_{n-1}^2}{b_n}$ go to $r_0$? Take the recurrence, and divide $b_{n-1}^2$ by both sides. This gives

$$ \frac{b_{n-1}^2}{b_n} = \frac{1}{3-2\left(\frac{b_{n-2}^2}{b_{n-1}}\right)^2},$$
or, letting $t_n = b_{n-1}^2/b_n$, 
$$t_n=\frac{1}{3-2 t_{n-1}^2}.$$
This has three fixed points, but the only attractive point is $r_0$, and $t_n$ approaches this point from any initial value except the two other fixed points.

Appendix:

Here is a sketch of the argument that if $\lim_{n\rightarrow \infty} t_n = r_0$, then $s$ exists. We'll consider an approximation to $s$ based on $b_n$. Let
$$s_n = \left(\frac{b_n}{r_0}\right)^{\frac{1}{2^n}}.$$ 
Then
$$\frac{s_n}{s_{n-1}} = \left(\frac{b_n}{r_0}\right)^{\frac{1}{2^n}}\left(\frac{r_0}{b_{n-1}}\right)^{\frac{2}{2^{n}}}
= \left( \frac{b_nr_0}{b_{n-1}^2}\right)^{\frac{1}{2^n}} 
= \left( \frac{r_0}{t_n}\right)^{\frac{1}{2^{n}}} .
$$
If $\lim_{n \rightarrow \infty} t_n = r_0$, this equation implies that $$ \prod_{m=n}^\infty \frac{s_{m+1}}{s_m} \rightarrow 1 \quad \mathrm{as} \quad n \rightarrow \infty,$$ and thus that $s =\lim_{n\rightarrow \infty} s_n $ exists. 

But in order to show that $b_n$ grows like $r_0 s^{2^n}$ as $n \rightarrow \infty$, you need not just that $s_n$ converges, but that it converges quickly. The exponent on $\frac{r_0}{t_n}$ above takes care of that. What we would like to show is that
$$\lim_{n\rightarrow\infty} \frac{s^{2^n}}{s_n^{2^n}} = 1.$$

But
$$
 \frac{s^{2^n}}{s_n^{2^n}}
= \prod_{m=n}^{\infty} \left(\frac{s_{m+1}}{s_m}\right)^{2^{n}}
= \prod_{m=n}^{\infty} \left(\frac{r_0}{t_{m+1}}\right)^{\frac{1}{2^{m+1}}2^{n}} \approx \left(\frac{r_0}{t_{n}}\right)^{2^n\left(\sum\limits_{m=n}^\infty \frac{1}{2^{m+1}}\right)}= \frac{r_0}{t_n},$$
which goes to $1$ as $n \rightarrow \infty$.