I have the following system: $\begin{cases} x = z^2 - y\\ z = y^2 - x\\ y = x^2 - z \end{cases}$
If we transform it by plugging in the variables at the end we get: $\begin{cases} x = z^2 - x^2 + z\\ z = y^2 - z^2 + y\\ y = x^2 - y^2 + x \end{cases}$
If we try the solution $(1, 1, 1)$ for the first and second system, we can notice, that new solutions are created and I am not sure why. Can somebody please explain?
You have the system of equations $$ \begin{cases} x=z^2-y\qquad(\text{eq}1)\\ z=y^2-x\qquad(\text{eq}2)\\ y=x^2-z\qquad(\text{eq}3) \end{cases} \qquad\qquad\qquad\qquad\qquad\;\;\; $$ As I noted in the comments, your proposed transformation yields a new system which is implied by the given system, but the two systems are not equivalent.
To solve the given system, we can proceed as follows . . .
I'll assume you only want real solutions.
Then we have \begin{align*} & y=x^2-z&&\bigl(\text{from }(\text{eq}3)\bigr) \;\;\;\;\, \\[4pt] \implies\;& y=x^2-(y^2-x)&&\bigl(\text{using }(\text{eq}2)\bigr) \\[4pt] \implies\;& x^2-y^2+x-y=0 \\[4pt] \implies\;& (x+y)(x-y)+(x-y)=0 \\[4pt] \implies\;& (x-y)(x+y+1)=0 \\[4pt] \implies\;& x=y\;\,\text{or}\;\,x+y=-1 \\[4pt] \end{align*} so we can consider two cases . . .
Case $(1)$:$\;x+y=-1$.
Then we get \begin{align*} & x=z^2-y&&\bigl(\text{from }(\text{eq}1)\bigr) \qquad\qquad\qquad\qquad\, \\[4pt] \implies\;& z^2=x+y \\[4pt] \implies\;& z^2=-1 \\[4pt] \end{align*} so there are no solutions for case $(1)$, since we are only considering real solutions.
Case $(2)$:$\;x=y$.
Then we get \begin{align*} & z=y^2-x&&\bigl(\text{from }(\text{eq}2)\bigr) \\[4pt] \implies\;& z=x^2-x \\[4pt] \implies\;& x=(x^2-x)^2-y&&\bigl(\text{using }(\text{eq}1)\bigr) \\[4pt] \implies\;& x=(x^2-x)^2-x \\[4pt] \implies\;& x^4-2x^3+x^2-2x=0 \\[4pt] \implies\;& x^3(x-2)+x(x-2)=0 \\[4pt] \implies\;& x(x-2)(x^2+1)=0 \\[4pt] \implies\;& x\in\{0,2\} \\[4pt] \implies\;& (x,y)\in\{(0,0),(2,2)\} \\[4pt] \implies\;& (x,y,z)\in\{(0,0,0),(2,2,2)\}&&\bigl(\text{using }(\text{eq}2)\bigr) \\[4pt] \end{align*} and it's easily verified that the triples above are valid solutions.