Solution:

Note first that $\sqrt{1+a_{n}^{2}+b_{n}^{2}}=3^{2^{n}}$. The proof is by induction; the base case follows trivially from what is given. For the inductive step, note that
$$1+a_{n+1}^{2}+b_{n+1}^{2}=1+a_{n}^{2}(1+a_{n}^{2}+b_{n}^{2})+b_{n}^{2}-2 a_{n} b_{n} \sqrt{1+a_{n}^{2}+b_{n}^{2}}+b_{n}^{2}(1+a_{n}^{2}+b_{n}^{2})+a_{n}^{2}+2 a_{n} b_{n} \sqrt{1+a_{n}^{2}+b_{n}^{2}}=1+(a_{n}^{2}+b_{n}^{2})(1+a_{n}^{2}+b_{n}^{2})+a_{n}^{2}+b_{n}^{2}=(1+a_{n}^{2}+b_{n}^{2})^{2}.$$
Invoking the inductive hypothesis, we see that $\sqrt{1+a_{n+1}^{2}+b_{n+1}^{2}}=(3^{2^{n}})^{2}=3^{2^{n+1}}$, as desired.

The quickest way to finish from here is to consider a sequence of complex numbers $\{z_{n}\}$ defined by $z_{n}=a_{n}+b_{n} i$ for all nonnegative integers $n$. It should be clear that $z_{0}=2+2 i$ and $z_{n+1}=z_{n}(3^{2^{n}}+i)$. Therefore,
$$z_{4}=(2+2 i)(3^{2^{0}}+i)(3^{2^{1}}+i)(3^{2^{2}}+i)(3^{2^{3}}+i).$$
This product is difficult to evaluate in the decimal number system, but in ternary the calculation is a cinch! To speed things up, we will use balanced ternary, in which the three digits allowed are $-1,0$, and $1$ rather than $0,1$, and $2$. Let $x+y i=(3^{2^{0}}+i)(3^{2^{1}}+i)(3^{2^{2}}+i)(3^{2^{3}}+i)$, and consider the balanced ternary representation of $x$ and $y$. For all $0 \leq j \leq 15$, let $x_{j}$ denote the digit in the $3^{j}$ place of $x$, let $y_{j}$ denote the digit in the $3^{j}$ place of $y$, and let $b(j)$ denote the number of ones in the binary representation of $j$. It should be clear that $x_{j}=-1$ if $b(j) \equiv 2\pmod{4}$, $x_{j}=0$ if $b(j) \equiv 1\pmod{2}$, and $x_{j}=1$ if $b(j) \equiv 0\pmod{4}$. Similarly, $y_{j}=-1$ if $b(j) \equiv 1\pmod{4}$, $y_{j}=0$ if $b(j) \equiv 0\pmod{2}$, and $y_{j}=1$ if $b(j) \equiv 3\pmod{4}$. Converting to ordinary ternary representation, we see that $x=221211221122001_{3}$ and $y=110022202212120_{3}$. It remains to note that $a_{4}=2x-2y$ and $b_{4}=2x+2y$ and perform the requisite arithmetic to arrive at the answer above.