5. $\{3,57\}$ is obtained from $\left\{\begin{array}{l}x+y+z=3 \\ x^{3}+y^{3}+z^{3}=3\end{array}\right.$ by eliminating $z$, resulting in $x^{3}+y^{3}+(3-x-y)^{3}=3$, which simplifies to $(x+y)^{2}-9(x+y)+9-xy(x+y)=1$, or $(x+y)(xy-x-y+9)=8$. Since $x, y, z \in \mathbf{Z}$, then $x+y= \pm 1, \pm 2$, $\pm 4$ or $\pm 8$. Solving this, we get $(x, y, z)=(1,1,1)$ or $(5,-4,-4)$ or $(-4,5,-4)$ or $(-4,-4,5)$, yielding $x^{2}+y^{2}+z^{2}=3$ or 57.
5. Given $x, y, z \in \mathbf{Z}$, such that $\left\{\begin{array}{l}x+y+z=3 \\ x^{3}+y^{3}+z^{3}=3\end{array}\right.$, then the set of all possible values of $x^{2}+y^{2}+z^{2}$ is . $\qquad$