Find solutions to $g : \mathbb{R} \to \mathbb{R}$ such that $g(x+y) = g(x) + g(y)$ and $g(x^z) = g(x)^z$ for $z \in \mathbb{R} \backslash \{0,1\}$. $z$ is a fixed number and not a variable.
Note : Please do not invoke any additional condition that forces $g$ to be linear
From the first condition, we need only two specializations: $y=x$ and $y=-x$, i.e., $$\tag 1g(2x)=2g(x)$$ and $$\tag2 g(x)+g(-x)=g(0).$$ From $(1)$ with $x=0$, we find $g(0)=0$ and then from $(2)$, $g(-x)=-g(x)$.
We will also use the second condition only where it is clear that it makes sense, i.e., when $x>0$ and it is known that $g(x)>0$.
From $g(x^2)=g(x)^2$, we see $g(x)=g(\sqrt x)^2\ge0$ for all $x\ge 0$. Assume $g$ is not identically $0$, i.e., $g(x)\ne0$ for some $x$ (so clearly $x\ne 0$). Then by $(1)$ and $(2)$ also $g(-x)\ne 0$ and $g(\pm2x)\ne0$. At least one of these four is a positive number $\ne1$. Thus we have $x_0\in(0,\infty)\setminus \{1\}$ with $g(x_0)>0$. Then for $x>0$, $x\notin\{1,x_0\}$, we have $$\ln g(x)=\ln g(x_0^{\log_{x_0}(x)})=\ln\left(g(x_0)^{\log_{x_0}(x)}\right)=\frac{\ln g(x_0)}{\ln x_0}\cdot\ln x.$$ In particular, $g(x)>0$ for all $x\notin \{-1,0,1\}$. If additionally $x\notin \{\frac12,\frac 12x_0\}$, we obtain $$\ln 2=\ln g(2x)-\ln g(x)=\frac{\ln g(x_0)}{\ln x_0}\cdot(\ln x+\ln 2)- \frac{\ln g(x_0)}{\ln x_0}\cdot\ln x=\frac{\ln g(x_0)}{\ln x_0}\cdot\ln 2$$ so that $$ g(x_0)=x_0.$$ But as $g(x)>0$ for all $x>0$ except possibly $1$ and $x_0$, we could have picked any positive number $\ne1$ for $x_0$. Thus we find $g(x)=x$ for all $x>0$ except possibly $x=1$. Then by $(1)$, also $g(1)=2g(\frac12)=1$ and by $(2)$, $g(x)=x$ for all $x$.