Answer: 983

Solution:

Write $a_n = 2^n \sin\theta_n$ with $\theta_0 = 0,$ so $\sqrt{4^n - a_n^2} = 2^n\,|\cos\theta_n|.$ Let $\theta = \arcsin\frac{3}{5},$ so $\cos\theta = \frac{4}{5}.$ The recursion becomes $$\begin{aligned} &a_{n+1} \\ &= 2^{n+1} \\ &\quad {}\cdot \left(\cos\theta \sin\theta_n + \sin\theta\,|\cos\theta_n|\right) \\ &= 2^{n+1}\sin(\theta_n \pm \theta), \end{aligned}$$ with the plus sign when $\cos\theta_n \ge 0$ and the minus sign when $\cos\theta_n \lt 0.$

Since $\frac{1}{2} \lt \frac{3}{5} \lt \frac{\sqrt{2}}{2},$ we have $30^\circ \lt \theta \lt 45^\circ.$ The angles $\theta,$ $2\theta$ have positive cosine, so the sequence of angles runs $0, \theta, 2\theta, 3\theta.$ But $90^\circ \lt 3\theta \lt 135^\circ$ has negative cosine, so $\theta_4 = 2\theta,$ and from then on the angle alternates between $3\theta$ and $2\theta.$ In particular $\theta_n = 2\theta$ for every even $n \ge 2.$

With $\sin 2\theta = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25},$ $$\begin{aligned} a_{10} &= 2^{10} \sin 2\theta \\ &= 1024 \cdot \frac{24}{25} \\ &= \frac{24576}{25} = 983.04, \end{aligned}$$ so the answer is $983.$