Fermat has stated that there are no natural numbers $x$, $y$, and $z$ such that $x^n + y^n = z^n$, in which $n$ is a natural number greater than $2$.
but here is where I have my question.
For example let's take $n = 12$, and write
$x^{12} + y^{12} = z^{12}$, isn't this the same as writing,
$(x^6)^2 + (y^6)^2 = (z^6)^2$, now that we can convert even powers to powers of $2$, won't there be a solution?
On
Write it as $$ (x^4)^3+(y^4)^3=(z^4)^3 $$ Indeed, the first reduction of FLT is to prove the statement for $n=4$ (which Fermat did) and for $n$ an odd prime.
Suppose $n>2$ and that $a^n+b^n=c^n$ for nonzero integers $a,b,c$. If $p$ is an odd prime such that $n=pm$, then we have $$ A^p+B^p=C^p $$ with $A=a^m,B=b^m,C=c^m$. If $n$ is not divisible by any odd prime, then it is a power of $2$ and, being $n>2$, we conclude $n=4k$, but then $$ A^4+B^4=C^4 $$ with $A=a^k,B=b^k,C=c^k$.
Thus, if we exclude the cases $n=4$ and $n$ an odd prime, we're done. The case $n=3$ was done by Euler, so FLT was proved for any multiple of $3$, including $n=12$.
The existence of $x, y, z$ such that $x^2+y^2=z^2$ indeed implies that
$$\left(\sqrt[6]{x}^6\right)^2 + \left(\sqrt[6]{y}^6\right)^2 = \left(\sqrt[6]{z}^6\right)^2$$
and therefore
$$\left(\sqrt[6]{x}\right)^{12} + \left(\sqrt[6]{y}\right)^{12} = \left(\sqrt[6]{z}\right)^{12}.$$
However, to show that this means $\sqrt[6]{x},\sqrt[6]{y},\sqrt[6]{z}$ indeed represent a solution (and therefore, a counterexample to Fermat's theorem), you would also need to show that the three numbers are integers.