First of all, it is equivalent to consider the objective function $\|AX\|_F^2$, which can be rewritten as
$$
\|AX\|_F^2 = \operatorname{tr}((AX)^T(AX)) =
\operatorname{tr}(X^TA^TAX) = \operatorname{tr}(A^TAXX^T)\\
= \operatorname{tr}[(A^TA)(XX^T)]
$$
To put this another way: in terms of the Frobenius inner product $\langle P,Q\rangle = \operatorname{tr}(P^TQ)$, we are looking to maximize $\langle A^TA, XX^T\rangle$.
To put this another way, we are given a positive semidefinite matrix $P_A = A^TA$ (which is generically of rank $m$ if $n > m$ and generically invertible otherwise), and we are looking for a positive semidefinite matrix $P_X = XX^T$ (which is generically of rank $k$ if $n > k$ and generically invertible otherwise). Our goal is to find the positive semidefinite matrix $P_X$ that maximizes $\operatorname{tr}(P_AP_X) = \langle P_A,P_X \rangle$ subject to the constraint that $\operatorname{tr}(P_X) = c$ and, in the case that $n > k$, the additional constraint that the rank of $P_X$ is at most equal to $k$. Notably, your suggested maximizer is of rank-1, so this constraint on the rank of $P_X$ is irrelevant.
From here, the problem is straightforward to solve. This inequality tells us that
$$
\operatorname{tr}(P_AP_X) = \langle P_A,P_X \rangle \leq \sigma_{\max}(P_A)\operatorname{Tr}(P_X) = \sigma_\max^2(A) \cdot c^2.
$$
From there, it suffices to show that if each column of $X$ is a multiple of $\mathbf v_1$, then the same holds for $P_{X}$, and it follows that
$$
P_AP_X = A^TA P_X = \sigma_\max^2(A) P_X,
$$
so that $\operatorname{tr}(P_AP_X) = c^2\sigma_\max^2(A)$ and the upper bound is attained. Thus, $X = \mathbf v_1\mathbf b^T$ indeed maximizes the objective function.
