why $x^{4} = 1 : x \in \mathbb{C}$ doesn't have infinite solutions?

All you have stated is that $$n+(n-1)i^2=n+(n-1)(-1)=1$$ $$n+(n+1)i^2=n+(n+1)(-1)=-1$$ has infinitely many solutions (which is trivially true). But if we were to solve the equation this way, we would need to substitute $x=a+bi$ in our original equation getting $$(a+bi)^4=a^4-6a^2b^2+b^4+(4a^3b-4ab^3)i=1$$ which (comparing real and imaginary parts) is equivalent to the simultaneous equations $$a^4-6a^2b^2+b^4=1$$ $$4a^3b-4ab^3=0$$ You should find this only has the solutions $(a,b)\in\{(1,0),(-1,0),(0,1),(0,-1)\}$. Hence there are only $4$ solutions for $x$ namely $$x=\pm1,\pm i$$

Actually, in $\mathbb C$ we have: $$x^4=1\Leftrightarrow x\in\sqrt[4]1,$$ where $$\sqrt[4]1=\{1,-1,i,-i\}$$

This is a good approach: Every non zero complex number can be determined by an angle and a length as follows: $$z=r(\cos\theta+i\sin\theta)~~~~~~~~~~~(\ast)$$ So, you need to find a number $z=x+iy$ (or a set of numbers) such that $$z^4=1=1+0\cdot i=\cos(2\pi)+i\cdot\sin(2\pi)$$ But $$cos(2\pi)=\cos(4\pi)=\cos(6\pi)=\cdots \mbox{ and } \sin(2\pi)=\sin(4\pi)=\sin(6\pi)=\dots$$ For this reason, the angle for our complex number will be only from $[0;2\pi\rangle$.

Then, as we said we need a lenght and a angle, in complex plane those properties are called module and argument:

If we need a number $z$ such that: $$z^4=1$$ it must satisfies $$\arg(z^4)=\arg(1)\mbox{ and } ||z^4||=||1||$$

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Finding the argument:

$\arg(z^4)=4\cdot\arg(z)$ but multiplying a number of the range $[0;2\pi\rangle$ four times may lead that it escapes from the interval $[0;2\pi\rangle$, so we need to add a correction factor of $2\pi$ several times ($k$ times) because $\cos,\sin$ has a period $2\pi$:$~~\arg(z^4)=4\cdot\arg(z)-2k\pi,~k\in\mathbb{Z}$

So we have: $$4\times\arg(z)-2k\pi = \arg(1)=0\Rightarrow\arg(z)=\frac{2k\pi}{4}$$

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Finding the module:

This is easiest:

$$||z^4||=||1||=1\Rightarrow||z||^4=1\Rightarrow ||z||=1 \mbox{ (normal real numbers root)}$$

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therefore, the complex $z$ number we are looking for, has a module $1$ ands its argument has to be of the form $\dfrac{2k\pi}{4}$. Since the argument must to be in the range $[0,2\pi\rangle$ we see that if $k>4\Rightarrow\dfrac{2k\pi}{4}>2\pi$ and since $\frac{2k\pi}{4}$ is increasing function respect $k$, the only values for $k$ are $0,1,2\mbox{ and }3$.

our numbers $z$ using $(\ast)$ are:

• $k=0: z_1=\cos(0)+i\sin(0)=1$
• $k=1: z_2=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})=0+1\cdot i=i$
• $k=2: \cos(\pi)+i\sin(\pi)=-1+0i=-1$
• $k=3: \cos(\frac{3\pi}{2})+i\sin(\frac{3\pi}{2})=0-1\cdot i=-i $

We conclude: If we are looking for all solutions for the equation $$z^n=w$$ then, those solutions are: $$z_k=\sqrt[n]{||w||}\cdot\left( \cos\left(\frac{\arg(w)+2k\pi}{n}\right) + i\cdot\sin\left(\frac{\arg(w)+2k\pi}{n}\right) \right),~k=0,1,\cdots,n-1$$