Is the proof I am using, sufficient/ correct for the system of equation?

Good job. Just too verbose.

You are correct into saying you can assume $x$, $y$ and $z$ all positive (there will be a corresponding solution with their negatives). The case when two are positive and one negative cannot appear, nor can the case of two negative and one positive, because the positivity of the left-hand sides forces positivity of the right-hand sides, so all three numbers must share the sign.

However, there is another simplification, namely, you can also assume $x$ is the maximum solution, because the equations are cyclic. Thus $$ x\ge y\ge z \qquad\text{or}\qquad x\ge z>y $$ You have already excluded the second case, so we can concentrate on the first.

In order to show that for a solution you need $x=y=z$, you just have to exclude $x>y$ and $y>z$.

In the case $x>y\ge z$, we have, according to your method, $$ x^4>y^4 \qquad y^2\ge z^2 $$ Then $$ 5yz=x^4+y^2+4>y^4+z^2+4=5zx $$ which implies $y>x$: a contradiction.

In the case $x\ge y>z$ we have $$ y^2>z^2 \qquad x^4\ge y^4 $$ which implies $$ 5yz=x^4+y^2+4>y^4+z^2+4=5zx $$ implying $y>x$, again a contradiction.

We saw that assuming either $x>y$ or $y>z$ leads to a contradiction. Since $x\ge y\ge z$ by assumption and we cannot have neither $x>y$ nor $y>z$, we deduce that $x=y$ and $y=z$.

Now, finding what's the common value is easy: we have $$ x^4-4x^2+4=0 $$ so $x^2=2$ and $x=\pm\sqrt{2}$. The problem has exactly two solutions.