The fact that each $T\in B(X,Y)$ is weakly compact does not imply $X$ or $Y$ reflexive. For example, every non-weakly compact operator $T:\ell_\infty\to Y$ is an isomorphism on a subspace isomorphic to $\ell_\infty$ (See Prop. 2.f.4 in Classical Banach spaces I, by Lindenstrauss ans Tzafriri).
Thus if $Y$ is a non-reflexive space containing no copy of $\ell_\infty$ (e.g. $\ell_1$, or a separable non-reflexive space) then every $T\in B(\ell_\infty,Y)$ is reflexive.