Choose a set $J$ and an $A$-module homorphism $\varphi : M \to \bigoplus_{j \in J} E$ which maps $M[\mathfrak m]$ isomorphically onto $(\bigoplus_{j \in J} E)[\mathfrak m] = \bigoplus_{j \in J} k$. We claim that $\varphi$ is an isomorphism, i.e., bijective. \medskip\noindent Injective. Let $z \in M$ be nonzero. Since $M$ is $\mathfrak m$-power torsion we can choose an element $f \in A$ such that $fz \in M[\mathfrak m]$ and $fz \not = 0$. Then $\varphi(fz) = f\varphi(z)$ is nonzero, hence $\varphi(z)$ is nonzero. \medskip\noindent Surjective. Recall that $F$ is flat, see Lemma \ref{lemma-frobenius-flat-regular}. Let $x_1, \ldots, x_d$ be a minimal system of generators of $\mathfrak m$. Denote $$ M_n = M[x_1^{p^n}, \ldots, x_d^{p^n}] $$ the submodule of $M$ consisting of elements killed by $x_1^{p^n}, \ldots, x_d^{p^n}$. So $M_0 = M[\mathfrak m]$ is a vector space over $k$. Also $M = \bigcup M_n$ by our assumption that $M$ is $\mathfrak m$-power torsion. Since $F^n$ is flat and $F^n(x_i) = x_i^{p^n}$ we have $$ M_n \cong (M \otimes_{A, F^n} A)[x_1^{p^n}, \ldots, x_d^{p^n}] = M[x_1, \ldots, x_d] \otimes_{A, F} A = M_0 \otimes_k A/(x_1^{p^n}, \ldots, x_d^{p^n}) $$ Thus $M_n$ is free over $A/(x_1^{p^n}, \ldots, x_d^{p^n})$. A computation shows that every element of $A/(x_1^{p^n}, \ldots, x_d^{p^n})$ annihilated by $x_1^{p^n - 1}$ is divisible by $x_1$; for example you can use that $A/(x_1^{p^n}, \ldots, x_d^{p^n}) \cong k[x_1, \ldots, x_d]/(x_1^{p^n}, \ldots, x_d^{p^n})$ by Algebra, Lemma \ref{algebra-lemma-regular-complete-containing-coefficient-field}. Thus the same is true for every element of $M_n$. Since every element of $M$ is in $M_n$ for all $n \gg 0$ and since every element of $M$ is killed by some power of $x_1$, we conclude that $M$ is $x_1$-divisible. \medskip\noindent Let $x = x_1$. Above we have seen that $M$ is $x$-divisible. If $\varphi$ is not surjective, then we can choose $e \in \bigoplus_{j \in J} E$ not in $M$. Arguing as above we may assume $\mathfrak m e \subset M$, in particular $x e \in M$. There exists an element $z_1 \in M$ with $x z_1 = x e$. Hence $x(z_1 - e) = 0$. Replacing $e$ by $e - z_1$ we may assume $e$ is annihilated by $x$. Thus it suffices to prove that $$ \varphi[x] : M[x] \longrightarrow \left(\bigoplus\nolimits_{j \in J} E\right)[x] = \bigoplus\nolimits_{j \in J} E[x] $$ is surjective. If $d = 1$, this is true by construction of $\varphi$. If $d > 1$, then we observe that $E[x]$ is the injective hull of the residue field of the regular ring $A/xA$, see Dualizing Complexes, Lemma \ref{dualizing-lemma-quotient}. Observe that $M[x]$ as a module over $A/xA$ is $\mathfrak m/(x)$-power torsion and we have \begin{align*} M[x] \otimes_{A/xA, F} A/xA & = M[x] \otimes_{A, F} A \otimes_A A/xA \\ & = (M \otimes_{A, F} A)[x^p] \otimes_A A/xA \\ & \cong M[x^p] \otimes_A A/xA \end{align*} Argue using flatness of $F$ as before. We claim that $M[x^p] \otimes_A A/xA \to M[x]$, $z \otimes 1 \mapsto x^{p - 1}z$ is an isomorphism. This can be seen by proving it for each of the modules $M_n$, $n > 0$ defined above where it follows by the same result for $A/(x_1^{p^n}, \ldots, x_d^{p^n})$ and $x = x_1$. Thus by induction on $\dim(A)$ we conclude that $\varphi[x]$ is surjective as desired.