What is wrong with the following chain of logic? Is it something to do with the fact that Arg is a set defined function?
$$z^2 = (-z)^2 \\\implies Arg((-z)^2) = Arg((-z)^2) \\\implies 2Arg(z)=2Arg(-z) \\\implies Arg(z)=Arg(-z) \\\implies Arg(1) = Arg(-1) $$
The problem is that $\arg(z^2)\not\equiv 2\arg(z)$. In fact $\arg(z^2)=2\arg(z)+2k\pi$, for some $k\in \{-1,0,+1\}$.
So from $\arg(z^2)=\arg((-z)^2)$, all you can conclude about $2\arg(z)$ and $2\arg(-z)$ is that they differ by a multiple of $2\pi$, since the values of $k$ which make the above equation true may be different for $z$ and $-z$.
The step $2Arg(z)=2Arg(-z) \implies Arg(z)=Arg(-z)$ is wrong. The possible values of $Arg$ should be considered elements in the set $\Bbb R$ modulo $2\pi$. This set is endowed with the usual addition operation, making it a group.
The notation $2Arg(z)$ is only a shorthand of $Arg(z)+Arg(z)$. You can see that there are elements $x,y$ in $\Bbb R$ modulo $2\pi$ distinct but $2x=2y$, for example $x=0$ and $y=\pi$. Thus, you cannot simply do division.
Note also that the equality $2Arg(z)=2Arg(-z)$ is meant to be understood as $2Arg(z)=2Arg(-z) \mod 2\pi$.
Your answer to your own question is correct. Each non-zero complex number has infinitely many arguments. In particular, the arguments of $1$ are the numbers of the form $k\pi$, with $k\in\mathbb Z$ and $k$ even, whereas the arguments of $-1$ are the numbers of the form $k\pi$, with $k\in\mathbb Z$ and $k$ odd. From this last statement you can deduce that every number of the form $k\pi$ with $k\in\mathbb Z$ and $k$ twice an odd number is an argument of $1(=(-1)^2)$. But it does not follow from this that these ar the only arguments of $1$.
$z^2 = (-z)^2$ is the point from which it all snowballs. It is of course true, but implications on the argument are to be handled with care for complex numbers. Writing a complex number $z$ as $z = r e^{i\theta}$, the equality, for example, $3^2 = (-3)^2$ in complex terms would translate to $$( 3 e^{i0} )^2 = 9 e^{i0} \,\,vs \,\, (3 e^{i\pi})^2 = 9 e^{2\pi i}$$ so you see, the arguments (both before eand after squaring) need not be the same (which is your conclusion), even if the moduli are. Arguments are defined modulo $2\pi$.
