If $a > b$, is $a^2 > b^2$?

If $\: \color{#c00}{a > b}\: $ then $\: a^2\! -\! b^2 = (\color{#c00}{a\!-\!b})(a\!+\!b) > 0 \iff a\!+\!b >0 $

no its not. When $a,b$ are positive it happens. Consider $a=-2$ and $b =-3$. notice that inequality reverses.

The correct statement is,$$|a|>|b|\iff a^2 > b^2 $$A counterexample of your hypothesis is $a = 7, b = -8.$

Yes, $a >b $, but $b^2 > a^2$, i.e.:$$ (-8)^2 > 7^2\\64 > 49$$