Any solvable group BS(1,m) is metabelian. A.H. Rhemtulla proved in (Commutators of certain finitely generated soluble groups. Canad. J. Math., 21 (1969), 1160-1164) that every finitely generated soluble of class $\leq 3$ group has a finite commutator width. Moreover,
P.W. Stroud (Topics in the theory of verbal subgroups.PhD Thesis, Univ. of Cambridge, 1966) proved that every verbal subgroup $w(G)$ of a finitely generated abelian-by-nilpotent group $G$ has a finite width. See for instance: D. Segal. Words: notes on verbal width in groups. London Math.Soc. Lect.Notes Ser. 361, Cambridge Univer. Press., 2009. I can add that the commutator width of the free metabelian group of rank $r \geq 2$ has the commutator width equal to $r.$ Hence every f.g. metabelian group has the commutator width $\leq r.$

Let $M$ be the free metabelian group with base $x, y$. Then the derived subgroup $M'$ is generated as a module over $Z[M/M']$ by $u = [x,y].$ Every element of $Z[M/M']$ can be written as $k + \alpha (1-x) + \beta (1-y).$ Then every element of $M'$ can be written as
$u^{k}[u^{\alpha },x][u^{\beta}, y]= [x,y^k] [u^{\gamma}, x][u^{\delta}, y].$ It gives bound 3. After some improving we get 2.