This is an expansion of the comment by Noam D. Elkies. Indeed, we can rewrite your equations 3) and 4) as $(\alpha,\beta)=(x^2,x^3)$ and $(\gamma,\delta)=(y^2,y^3)$ for some $x,y$. Substituting into eqs. 1) and 2) these expressions of $\alpha,\beta,\gamma,\delta$ in terms of $x,y$, we reduce your system of four eqs. to that of the following two: $P_1=0=P_2$, where $P_1,P_2$ are certain polynomials in $x,y$.
The resultant $\text{Res}_y(P_1,P_2)$ vanishes iff, for a given value of $x$, the polynomials $P_1,P_2$ have a common root $y$. We find that $\text{Res}_y(P_1,P_2)=0=\text{Res}_x(P_1,P_2)$ identically for all $x$ and $y$. So, $P_1$ and $P_2$ need some preliminary cleaning. Indeed, we see that
\begin{equation}
Q_1:=\frac{P_1}{y-x},\quad Q_2:=\frac{P_2}{2(y-x)}
\end{equation}
are polynomials in $x,y$. Thus,
\begin{equation}
\text{$(x,x)$ is a solution to our system for any complex $x$.} \tag{*}
\end{equation}
The resultants $R_y:=\text{Res}_y(Q_1,Q_2)$ and $R_x:=\text{Res}_x(Q_1,Q_2)$ are polynomials in $x$ and in $y$, respectively, each having $16$ distinct complex roots, say $x_1,\dots,x_{16}$ for $R_y$ and $y_1,\dots,y_{16}$ for $R_x$. Details of all calculations here, as well as the particular enumeration of $x_1,\dots,x_{16}$ and $y_1,\dots,y_{16}$, can be seen in the Mathematica notebook or its pdf image.
So, for each $i\in[16]:=\{1,\dots,16\}$ there is at least one $j\in[16]$ such that the pair $(x_i,y_j)$ is a solution to the system $Q_1=0=Q_2$. Moreover, it is easy to see that for each $i\in[16]\setminus\{3\}$, there is at most one (and therefore the only one) $j=j_i\in[16]$ such that the pair $(x_i,y_{j_i})$ is a solution to the system $Q_1=0=Q_2$. The corresponding pairs $(i,j_i)$ are $(1, 3), (2, 4), (4, 2), (5, 6), (6, 5), (7, 11), (8, 12), (9, 10), (10, 9), (11, 7)$, $(12, 8), (13, 14), (14, 13), (15, 16), (16, 15)$. For the exceptional value $i=3$, we have at most two values of $j\in[16]$ (namely, $j=1$ and $j=3$) such that the pairs $(x_3,y_j)$ may be solutions to the system $Q_1=0=Q_2$; in fact we see that these pairs $(x_3,y_j)$, equal $(0,\sqrt3)$ and $(0,0)$, are indeed solutions to the system $Q_1=0=Q_2$; the pair $(0,0)$ has already been accounted for by (*).
Thus, we have described all the $16$ pairs $(x,y)$ that are (in addition to the "trivial" pairs given by (*)) solutions to the system $Q_1=0=Q_2$. All these solutions $(x,y)$ are now straightforward to transcribe into a complete set of solutions $(\alpha,\beta,\gamma,\delta)$ of the original system.
