Let $\mathbb{F}_2$ be the free group on two generators. Does $\mathbb{F}_2 \times \mathbb{F}_2$ embed as a subgroup of $\mathrm{SL}_3(\mathbb{Z})$?

2$\begingroup$ No; more generally, let $G$ be a nonvirtuallysolvable subgroup of $\mathrm{GL}_3(\mathbf{C})$. Then the centralizer of $G$ is abelian. Indeed its Zariski closure contains a Zariskiclosed copy of $\mathrm{(P)SL}_2(\mathbf{C})$. There are two such subgroups up to conjugation: the irreducible $\mathrm{PSL}_2(\mathbf{C})(=\mathrm{SO}_3)$ and the upperleft block. The first has a trivial centralizer, and the second has centralizer equal to the diagonal matrices $(a,a,b)$. [I'm pretty sure this argument already exists somewhere on this site.] $\endgroup$ – YCor Jan 8 '19 at 9:26
Here is a brief elementary argument only relying on Burnside's theorem that proper unital subalgebras of matrix algebras are nonirreducible (over an algebraically closed field).
Proposition: Let $K$ be an algebraically closed field. Let $A,B$ be noncommuting matrices in $\mathrm{M}_3(K)$. Then the centralizer of $\{A,B\}$ in $\mathrm{M}_3(K)$ is triangulable, i.e., stabilizes a flag (i.e., is conjugate into the subalgebra of upper triangular matrices).
Proof: there are essentially 6 types of $3\times 3$ matrices: (a) 3 eigenvalues (b) 2 eigenvalues, diagonalizable (c) 2 eigenvalues, not diagonalizable (d) scalar (=1 eigenvalue, diagonalizable) (e) 1 eigenvalue, scalar+(nilpotent of rank 1) (f) 1 eigenvalue, scalar+(nilpotent of rank 2).
Matrices of type (a),(c),(f) have abelian centralizer. Matrices of type (e) have triangulable centralizer (hint: compute the centralizer of the matrix $E_{13}$). By assumption, $A$ is not central and not of type (d). If $A$ has type (acef) then it has triangulable centralizer. So the only case to consider is when $A$ has type (b): we can suppose that $A$ is the diagonal matrix $(0,0,1)$. The centralizer $C$ of $A$ is the set of matrices diagonal by blocks $2+1$, and the double centralizer $C'$ is reduced to $K+KA$ (so $C'\subset C$). Hence $B\notin C$. Therefore, the intersection $I$ of the centralizers of $A$ and $B$ is properly contained in $C$, and hence by the contraposite of Burnside's theorem (in dimension 2, in which it is an elementary exercise) implies that $I$ is triangulable.
Corollary: for every field $K$, every nonabelian subgroup of $\mathrm{GL}_3(K)$ has a 3step solvable centralizer. In particular, no group of the form $H_1\times H_2$ with $H$ nonabelian and $H_2$ nonsolvable, is embeddable into $\mathrm{GL}_3(K)$ for any field $K$.
Remarks:
1) a slight refinement shows that a 3step solvable subgroup has an abelian centralizer, so "3step" can be replaced with "2step").
2) Here are two commuting nonabelian subgroups of $\mathrm{SL}_3(\mathbf{Q})$, each isomorphic to the BaumslagSolitar $\mathrm{BS}(1,p^3)$ (here $p\in\mathbf{Z}\smallsetminus\{0,1\}$), with trivial intersection, thus generating their direct product: $$\Gamma_1=\left\langle\begin{pmatrix}p & 0 & 0\\0&p^{2}&0\\0&0&p\end{pmatrix},\begin{pmatrix}1 & 1 & 0\\0&1&0\\0&0&1\end{pmatrix}\right\rangle,\;\Gamma_2=\left\langle\begin{pmatrix}p & 0 & 0\\0&p&0\\0&0&p^{2}\end{pmatrix},\begin{pmatrix}1 & 0 & 1\\0&1&0\\0&0&1\end{pmatrix}\right\rangle.$$
This cannot exist. See Misha Kapovich's answer to this question: https://mathoverflow.net/a/163754/1345
In particular, part (2) of his answer implies that there can be no semisimple $\mathbb{F}_2\times \mathbb{F}_2$ subgroup of $SL_3(\mathbb{Z})$. That is, there is no such subgroup in which every element is diagonalizable.
Now one sees that if there is an arbitrary $\mathbb{F}_2\times \mathbb{F}_2$ subgroup of $SL_3(\mathbb{Z})$, then each $\mathbb{F}_2$ factor contains a semisimple $\mathbb{F}_2$ subgroup. Passing to these subgroups, we obtain a semisimple such subgroup, a contradiction.

$\begingroup$ What's a reference for the fact that any $F_2$ contains a semisimple $F_2$? $\endgroup$ – YCor Jan 9 '19 at 0:36

$\begingroup$ @YCor: This follows from some version of the proof of Tits' alternative, but I don't have a specific reference. Actually, it suffices just to see that there is a single element with 3 distinct eigenvalues (this is semisimple). If not, one can see that each matrix is quasiunipotent (with eigenvalues $\pm1$). Then one can see that the subgroup is virtually nilpotent, contradicting that it is free. Once one has such an element with 3 distinct eigenvalues in one factor, then all the elements in the other factor must have the same eigenspaces in order to commute. But this is a contradiction. $\endgroup$ – Ian Agol Jan 9 '19 at 2:28

2$\begingroup$ OK. Well if I can put it otherwise, you prove that if $G$ is a nonvirtuallyunipotent subgroup of $\mathrm{GL}_3(\mathbf{Z})$ then its centralizer is abelian semisimple [this conclusion fails in $\mathrm{GL}_3(\mathbf{Q})$ as I have noticed in my answer]. Indeed, considering the Zariski closure of $G$, one sees that $G$ has an element $g$ with eigenvalues not contained in $\{\pm 1\}$, and hence (using that $g$ is integral) $g$ has 3 distinct eigenvalues, hence its centralizer is abelian. $\endgroup$ – YCor Jan 9 '19 at 9:39