The finite subgroups of $GL_3(\mathbb{Z})=PGL_3(\mathbb{Z})$ are known in the literature:

$\qquad$ Tahara: On the finite subgroups of $GL(3,\mathbb{Z})$. Nagoya Math. J. 41(1971), 169-209.

In particular Proposition 3 states that there are exactly two non-conjugate subgroups of order three. Representants are $$\langle \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & -1 \end{pmatrix}\rangle, \qquad \begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}\rangle$$

The finite subgroups of $GL_3(\mathbb{Z})$ are known in the literature:

$\qquad$ Tahara: On the finite subgroups of $GL(3,\mathbb{Z})$. Nagoya Math. J. 41(1971), 169-209.

In particular Proposition 3 states that there are exactly two non-conjugate subgroups of order three. Representants are $$U_1=\langle \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & -1 \end{pmatrix}\rangle, \qquad U_2=\langle\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}\rangle$$

Added: In $PGL_3(\mathbb{Z})=GL_3(\mathbb{Z})/\langle -I\rangle$ there are also exactly two conjugacy classes of subgroups of order 3 and representants are $\bar{U}_1,\bar{U}_2$.

For, let $V_i=\langle x_i\rangle \le G := GL_3(\mathbb{Z}),i=1,2$ be subgroups of order 3. If $\bar{V}_i$ are conjugate in $\bar{G} :=PGL_3(\mathbb{Z})$ then there is $g \in G$ s.t. $x_2=(\pm I)gx_1^kg^{-1} (k=1,2)$ and hence $(\pm I)^3=I$ and $x_2=gx_1^kg^{-1}$, i.e. the $V_i$ are conjugated in $G$.

Conversely, let $\bar{V}$ be a subgroup of $\bar{G}$ of order three. It's preimage in $G$ has order 6. Hence there is $V \le G$ of order three that maps to $\bar{V}$ and by the above $V$ is conjugated to some $U_i$.