Prove that if $\mathcal{N}(A) \subseteq \mathcal{N}(B)$ then $\mathcal{R}(B^\intercal) \subseteq \mathcal{R}(A^\intercal)$

It was asked to prove the following statement:

$ \mathcal{N}(A) \subseteq \mathcal{N}(B) \rightarrow \mathcal{R}(B^\intercal) \subseteq \mathcal{R}(A^\intercal)$

For me, it is intuitively true (somewhat because of the rank theorem), and I want to prove by contradiction. So I am supposing a $ z \in \mathcal{R}(B), \not\in \mathcal{R}(A)$ , but I can’t go on beyond this. I am always rounding the fact that $ z$ is a linear combination of the rows of $ B$ , that the null space and the row space are both founded in the echelon form (though the former is related to the reduced echelon form), but that is it. What additionals properties do $ z$ must have so we can get a contradiction?