Is there a criterion for which $BS(m,n)$ are solvable (and non-solvable)? If not, are there classes of such groups where this is known? Let $BS(m,n) = \langle a,t\mid ta^mt^{-1} = a^n \rangle$ be a Baumslag-Solitar group, with $m,n \in \mathbb{Z}.$

Is there a criterion for which $BS(m,n)$ are solvable (and non-solvable)? If not, are there classes of such groups where this is known?

It is already clear to me that $BS(1,n)$ are all solvable since they are the semidirect product of abelian groups.
This question arose from identifying whether or not $BS(2,3)$ and $BS(2,4)$ are solvable, which at the very least I would greatly appreciate in learning the answer to.
 A: 
It's solvable iff $1\in\{|m|,|n|\}$.

As you said, if $1\in\{|m|,|n|\}$ then you get a semidirect product, say $\mathrm{BS}(1,n)=\mathbf{Z}[1/n]\rtimes\mathbf{Z}$ for $n\neq 0$ and $=\mathbf{Z}$ for $n=0$.
Otherwise this group has a free non-abelian subgroup, and indeed its second derived subgroup is a free group (so it's residually solvable anyway).
In the cases $|m|=|n|\ge 2$ or ($mn=0$ and $\max(|m|,|n|)\ge 2$) the group is actually virtually free non-abelian. ($\mathrm{BS}(0,n)\simeq\mathbf{Z}\ast C_{|n|}$). In the remaining cases (namely $|m|\neq |n|$ and $\min(|m|,|n|)\ge 2$ it is known not to be residually finite.
A standard way to show that they're indeed non-abelian is to use that these are non-ascending HNN extensions (HNN extension by a group over an isomorphism between two proper subgroups): each such group, by Bass-Serre theory, acts on a regular tree of valency $\ge 3$ with no invariant proper subtree and no fixed point at infinity, and this forces the existence of a non-abelian free subgroup.
