Is there (or rather what is) a more elegant/efficient way to solve the DLP for $g^x=h$ in $GL_3(\mathbb{F}_2)$ where
$$g=\begin{pmatrix}0 &1 & 1 \\ 1 &1 &1 \\ 1&0&1 \end{pmatrix}$$ and
$$h=\begin{pmatrix}1 &1 & 0 \\ 0 &1 &1 \\ 1&1&1 \end{pmatrix}$$ other than using the brute force method which will reveal the answer to be $x=6+7m$?
The characteristic polynomial of $g$ over $\mathbb F_2$ is $p(t) = t^3 + t + 1$ which is irreducible. The splitting field of this polynomial has $2^3 = 8$ elements, so any nonzero element of it satisfies $r^7 = 1$. The eigenvalues of $g^7$ are all $1$, and $g^7$ must be diagonalizable over the splitting field since $g$ is, so we must have $g^7 = I$.
Now if $h = g^k$, an eigenvector of $g$ for eigenvalue $r$ will be an eigenvector of $h$ corresponding for eigenvalue $r^k$. An eigenvector of $g$ for eigenvalue $r$ is $$ v = \pmatrix{ r\cr r+1\cr r^2 + r + 1\cr}$$ and $$ h v = \pmatrix{1 \cr r^2 \cr r^2 + r\cr} = r^{-1} v$$ Thus $h = g^{-1} = g^{6 + 7 m}$ for any $m$.