What is wrong with my 'proof' of $i=1$?

This has nothing to do with complex numbers. By the same argument, $-1=1$, since$$(-1)^2=1^2\implies\sqrt{(-1)^2}=\sqrt{1^2}\implies-1=1.$$The error lies in assuming that $\sqrt{x^2}=x$. Actually, $\sqrt{x^2}=|x|$. In the case of complex numbers, it's even worst, since every complex number (other than $0$) has four fourth roots. So, the expression $\sqrt[4]z$ doesn't make sense unless and until you decide which fourth root of $z$ you have in mind. Even then, it will often be false that $\sqrt[4]{z^4}=z$.

Square roots don't work like that. With the same argument you used, you could have done $$ (-1)^2=1^2\ \ \ \implies\ \ \ -1=1. $$ It has nothing to do with complex numbers.

If this may reconcile you with complex numbers, consider this: a complex number is a pair of reals, and we define

• addition as $(a,b)+(c,d):=(a+b,c+d)$,

• multiplication as $(a,b)(c,d):=(ac-bd,ad+bc)$.

Then we accept the "shorthand" conventions $1=(1,0)$ and $i=(0,1)$. It is an easy matter to show that these rules form a consistent arithmetic with the four basic operation, where you can freely use these equivalences.

We also have the consequences

$$i^2=(0,1)(0,1)=(-1,0)=-1$$

and

$$i^4=(-1,0)(-1,0)=(1,0)=1.$$

But that means in no way that $i=1$. It just means that $i$ is one of the solutions of the equation

$$z^4=(1,0)=1.$$

All solutions are (see below)

$$(1,0)=1,\\(0,1)=i,\\(-1,0)=-1,\\(0,-1)=-i.$$

Unless you specify a convention, in the complex we don't know which solution $$\sqrt[4]1$$ denotes.

Final remark: in this discussion, we only used real numbers and the special symbol $i$.

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$$z^4=(x,y)^4=(x^4-6x^2y^2+y^4,4yx^3-4xy^3)=(1,0)$$ requires $$x=0\lor y=0\lor x^2=y^2$$

(by cancelling the imaginary part).

By plugging these in the real part, we get $$y^4=1\lor x^4=1\lor -4x^4=-4y^4=1.$$

As $x,y$ are real, the only options are

$$(\pm1,0),(0,\pm1).$$