What is awry with this proof? Let $x=5$, $y=7$, $z=6$
$x+y = 2z$
Rearranging, $x-2z = -y$
and $x = -y+2z$
Multiply both sides respectively. $x^2-2xz = y^2-2yz$
$$x^2-2xz+z^2 = y^2-2yz+z^2$$
$$(x-z)^2 = (y-z)^2$$
$$x-z = y-z$$
Hence $x=y$, or $5 = 7$
Well, the conclusion is clearly false, but what went wrong? I think it may be the step in which one square roots both sides because it's taking out one solution?

You have $x=−y+2z$ and $x−2z=−y$  then you multiply both sides.
I think you are computing $-xy$ in two ways: for the first $-xy=y^2-2zy$
for the second $-xy=-2zx+x^2$ so you have $y^2-2zy=x^2-2zx$.  Adding $z^2$ to both sides
gives $(y-z)^2=(x-z)^2$. But now taking square roots gives $|y-z|=|x-z|$. 
If you try to remove the absolute value signs then you get $y-z=x-z$ if both are positive or negative but $y-z=-(x-z)$ if these have opposite signs. In the first case this leads to $y=x$, and in the second case $y=-x+2z$. For your example, the second case applies
$7=-5+2\cdot6$. 
