This question does not make sense to me. For one I believe it does not tell us if $\gamma$ is in our field $E$. For all we know $\gamma$ and for that matter $\xi$ can have a degree of $1$ or $2$.
Here is the problem:
let $a_1, a_2 \in E$ and $\gamma, \xi \in \bar{E} \ s.t. \ \gamma^2= a_1, \ \xi^2= a_2 $ show that: $E(\gamma) = E(\xi) \iff \exists \ a_3 \in E \ s.t. a_1 = a_3^2 a_2 $
I tried to look at this in terms of roots of $x^2 - 2x -1 $ but $\nexists a_3 \ s.t. (1-\sqrt2)^2 = a_3^2(1+\sqrt2)^2 $
Perhaps you are confusing equality of fields with isomorphism. The problem asks about the former, whereas your example, if I understand it correctly, has to do with the latter.
Suppose that $\gamma^2 = a_1 \in E \smallsetminus \{0\}$. Then either $\gamma \in E$, in which case $E(\gamma) = E$, or $\gamma \not\in E$, in which case $E(\gamma) = E \oplus E\gamma$ is two-dimensional over $E$. Supposing now that $\xi^2 = a_2 \neq 0$ and $E(\gamma) = E(\xi)$, we see that if $\gamma \in E$, then also $\xi \in E$, and so $a_1 / a_2 = (\gamma/\xi)^2$ is a square in $E$. For the more interesting case, suppose that $\gamma \not \in E$. Then neither is $\xi$. But $\gamma \in E(\xi)$, since $E(\gamma) = E(\xi)$ so we can write $\gamma = b + c\xi$. Squaring gives $a_1 = b^2 + c^2 a_2 + 2bc\xi$. Since the left hand side is in $E$, so too must the right-hand side be; therefore $2bc=0$. Let's come back to the case when the characteristic is $2$; if it is not, then either $b$ or $c$ must be $0$, but it cannot by $b$, since $\gamma \not\in E$. So $\gamma = c\xi$, so $a_1/a_2 = (\gamma/\xi)^2 = c^2$ is a square in $E$.
In the converse direction, if $a_1/a_2 = (\gamma/\xi)^2 = c^2$ is a square in $E$, then $E + E\gamma = E + Ec\xi = E+E\xi$, and $E(\gamma) = E(\xi)$.
Finally, there is the issue of dividing by $2$. A finite field of characteristic $2$ has multiplicative group of abelian odd order (in fact, cyclic), and so squaring is bijective. (And if $a_1 = d^2$ is a square in $E$, then $(d - \gamma)(d+\gamma) = 0$, so $\gamma = \pm d$.) But there are infinite fields of characteristic $2$ that contain non-squares; the fraction field $\mathbb F_2(x)$, for example.
Consider, then, $E = \mathbb F_2(x)$, with $\gamma = \sqrt x$ and $\xi = 1 + \sqrt x$. Note that $\xi^2 = 1 + x \in E$. But $(1+x )/x = 1 + 1/x$ is not a square in $E$. Thus the problem should include the condition that the characteristic is not $2$.