Let $i = \sqrt{-1}$ and $x \in \mathbb{R}$, then we know that $x^2\geqslant0$ for all $x \in\mathbb R$. Therefore, $$x^2 > -1$$ which implies $x > i$? Hence, every real number is greater than $i$ since $x∈ \mathbb R$. Now, this implies, $i<0$. Hence, $i$ is negative.
Now, since $i$ is negative, let $i = -m$ where $m$ is the magnitude of $i$.
We know that $i = \sqrt{-1}$. Substituting the value of $m$, we get: $-m = \sqrt{-1}$.
Squaring both the sides, we get: $$m^2= -1 \implies m = \sqrt{-1}$$ Hence, $$i = -m = -i$$ $$⇒i = -i ⇒i = 0$$
This is a contradiction to (i) that $i<0$ as at the same time, $i=0$
Then, what is $i$, or what is $ \sqrt{-1}$ exactly? What does this contradiction prove?
To summarize, there are many problematic assumptions and missteps in reasoning:
As pointed out in the comments, $x^2>-1$ does not imply $x>i$. The square root function is increasing over non-negative reals. Also, statements saying one complex number is greater than another (when either of them is not real) are meaningless.
NB: As also pointed out in the comments, the complex numbers cannot be made into an ordered field under usual addition and multiplication. See here if interested in further discussion on ordering complex numbers.
If for $x,a \in \mathbb{R}$, $x^2\geq a\geq 0$, we would conclude either $x\geq \sqrt{a}$ or $x\leq -\sqrt{a}.$
If you did happen to show that every real number is greater than "something," that "something" would certainly not be a finite negative quantity (consider the real number that is one less than that quantity for a contradiction).
Independent of the above issues, what ultimately gives the fallacious result $i=0$ is this misstep: $$-m=\sqrt{-1}\implies m^2=-1\implies m=\sqrt{-1}.$$ Instead, if $m^2=a$, we would conclude either $m=\sqrt{a}$ or $m=-\sqrt{a}$ .