Preliminary Analysis:
$$|x^2-4| = |(x-2)(x+2)| = |x-2|<\frac{\varepsilon}{|x+2|}$$
Since $\delta$ can not depend on $x$ then
choose a small $\delta_i$ such that $$\delta_i<\frac{\varepsilon}{|x+2|}$$ where $\delta_i$ does not depend on $x$
Impose $|x-2|≤ 1$ then
$$-1 ≤x-2 ≥1$$
$$ = 1 ≤x ≥3$$
$$= 3 ≤x+2 ≥5$$
$$ = -5<3 ≤x+2 ≥5$$
$$ = |x+2| ≤5 $$
put $\delta_i = \frac{\varepsilon}{5}$
then $$\delta_i = \frac{\varepsilon}{5}<\frac{\varepsilon}{|x+2|}$$
Now we have $$|x^2-4| = |(x-2)(x+2)| < 5.|x-2| = |x-2|<\frac{\varepsilon}{5}$$
Formal proof:
$\lim_{x\to 2}f(x)=x^2\iff (\forall\varepsilon>0)(\exists\delta>0):0<\lvert x-2\rvert<\delta = \min({1,\frac{\varepsilon}{5}})\implies\bigl\lvert x^2-4\rvert = |(x+2)(x-2)| < 5.|x-2|<5 \frac{\varepsilon}{5} = \varepsilon$
We want to show that $\lim_{x \to 2} x^2 = 4$.
We know that if $\lim_{x \to a} f(x) = L$, then for each $\epsilon > 0$, there exists $\delta > 0$ such that whenever $0 < |x - a| < \delta$, then $|f(x) - L| < \epsilon$.
In this case, we claim $L = 4$, so given $\epsilon > 0$, we wish to find $\delta > 0$ such that whenever $0 < |x - 2| < \delta$, $|x^2 - 4| < \epsilon$.
Let $\epsilon > 0$. Observe that $$|x^2 - 4| = |(x + 2)(x - 2)| = |x + 2||x - 2|$$ Let $\delta_1 = 1$. If $0 < |x - 2| < 1$, then $$-1 < x - 2 < 1 \implies 1 < x < 3 \implies 3 < x + 2 < 5 \implies |x + 2| < 5$$ Set $\delta_2 = \frac{\epsilon}{5}$. Let $\delta = \min\{\delta_1, \delta_2\}$. Then if $0 < |x - 2| < \delta$, $$|x^2 - 4| = |x + 2||x - 2| < 5|x - 2| < 5 \cdot \frac{\epsilon}{5} < \epsilon$$ so we may conclude that $\lim_{x \to 2} x^2 = 4$.
Note: You made some errors in your preliminary analysis.
Your claim in the first line that $|(x - 2)(x + 2)| = |x - 2|$ is only true when $|x + 2| = 1$. It looks like you were attempting to isolate $|x - 2|$ without paying attention to what your equals signs meant.
If $|x - 2| \leq 1$, then $-1 \leq x - 2 \color{red}{\leq} 1$. You reversed the direction of the inequality in that line and the following three lines.
Also, you should not write $\delta_i$ for two different values of $\delta$. Name each one separately, then take their minimum to ensure that all the bounds are satisfied.