Field with 2 elements Let $ A $ be a ring with $ x^{2}+y^{2}+z^{2}=xy+yz+zx+xyz+1,\forall x,y,z\in A^{*} $. 
Prove that $ A $ is a field with 2 elements.

If we put $ x=y=z=1 $ we obtain that $ 1+1=0 $. 
If we put $y=z=1 $ we have that $ x^2=x $, which means $ A $ is a boolean ring.
That's all I did so far.
Hint: Try setting $x=y=z$.

(By the way, it is not correct to conclude that $|A|=2^n$ as you have done, since $A$ might be infinite.  Also, the statement you are trying to prove is slightly incorrect, since the condition also vacuously holds if $A=\{0\}$.  Your conclusion should be that $A$ is either $\{0\}$ or a field with two elements.)
$x^2 + x^2 + x^2 = x*x + x*x + x*x + x*x *x + 1 =x^2 + x^2 +x^2 + x^3 + 1$ so $x^3 + 1 = 0$ and $x^3 = -1$

$x^2 + x^2 + 1 = x*x + x*1 + 1*x + x*x *1 = x^2 + x + x + x^2 + 1$ so  $x + x =0$ and$x = -x $ and therefore  $x^3 = -1 = 1$

$x^2 = x^2 + ( 1-1) = x^2 + 1 + 1 = x^2 + 1^2 + 1^2 = x*1 + 1*1 + 1*x + x*1*1 + 1=x+x + x +1 + 1 = x$

So $1=x^3 = x(x^2) = xx = x^2 = x$

So $x =1$ for all $x \ne 0$.

Any ring with only two elements is a field.