I think Grushko plus the Freiheitssatz does the trick. Suppose that $G=A\ast B$ is a 1-relator group which splits as a free product non-trivially. By Grushko, $rank(G)=rank(A)+rank(B)=m+n$, where $rank(A)=m, rank(B)=n$. If $G$ is not free, then by Grushko there is a 1-relator presentation $\langle x_1,\ldots, x_m ,y_1,\ldots, y_n | R\rangle$, such that $\langle x_1,\ldots, x_m \rangle =A \leq G, \langle y_1, \ldots, y_n \rangle=B \leq G$ (for this, one has to use the strong version of Grushko that any 1-relator presentation is Nielsen equivalent to one of this type). Suppose that $R$ is cyclically reduced, and involves a generator $x_1 \in F_m$. Then $\langle x_2,\ldots , x_m, y_1, \ldots, y_n\rangle$ generates a free subgroup of $G$ by the Freiheitssatz. But this implies that $B = \langle y_, \ldots, y_n\rangle$ is free. Moreover, if $R$ involves one of the generators $y_i$, then one sees that $A=\langle x_1,\ldots,x_m\rangle$ is also free, and therefore $G=A\ast B$ is free, a contradiction. So $R \in F_m$, as required.

$\newcommand{\rank}{\operatorname{rank}}$I think Grushko plus the Freiheitssatz does the trick. 

Suppose that $G=A \ast B$ is a one-relator group which splits as a free product non-trivially. By Grushko, $\rank(G)=\rank(A)+\rank(B)=m+n$, where $\rank(A)=m$ and $\rank(B)=n$. If $G$ is not free, then by Grushko there is a one-relator presentation $\langle x_1,\ldots, x_m ,y_1,\ldots, y_n | R\rangle$, such that $\langle x_1,\ldots, x_m \rangle =A \leq G, \langle y_1, \ldots, y_n \rangle=B \leq G$ (for this, one has to use the strong version of Grushko that any one-relator presentation is Nielsen equivalent to one of this type). Suppose that $R$ is cyclically reduced, and involves a generator $x_1 \in F_m$. Then $\langle x_2,\ldots , x_m, y_1, \ldots, y_n\rangle$ generates a free subgroup of $G$ by the Freiheitssatz. But this implies that $B = \langle y_, \ldots, y_n\rangle$ is free. Moreover, if $R$ involves one of the generators $y_i$, then one sees that $A=\langle x_1,\ldots,x_m\rangle$ is also free, and therefore $G=A\ast B$ is free, a contradiction. So $R \in F_m$, as required.

I think Grushko plus the Freiheitssatz does the trick. Suppose that $G=A\ast B$ is a 1-relator group which splits as a free product non-trivially. By Grushko, $rank(G)=rank(A)+rank(B)=m+n$, where $rank(A)=m, rank(B)=n$. If $G$ is not free, then by Grushko there is a 1-relator presentation $\langle x_1,\ldots, x_m ,y_1,\ldots, y_n | R\rangle$, such that $\langle x_1,\ldots, x_m \rangle =A \leq G, \langle y_1, \ldots, y_n \rangle=B \leq G$ (for this, one has to use the strong version of Grushko that any 1-relator presentation is Nielsen equivalent to one of this type). Suppose that $R$ is cyclically reduced, and involves a generator $x_1 \in F_m$. Then $\langle x_2,\ldots , x_m, y_1, \ldots, y_n\rangle$ generates a free subgroup of $G$ by the Freiheitssatz. But this implies that $B = \langle y_, \ldots, y_n\rangle$ is free. Moreover, if $G$ involves one of the generators $y_i$, then one sees that $A=\langle x_1,\ldots,x_m\rangle$ is also free, and therefore $G=A\ast B$ is free, a contradiction. So $R \in F_m$, as required.

I think Grushko plus the Freiheitssatz does the trick. Suppose that $G=A\ast B$ is a 1-relator group which splits as a free product non-trivially. By Grushko, $rank(G)=rank(A)+rank(B)=m+n$, where $rank(A)=m, rank(B)=n$. If $G$ is not free, then by Grushko there is a 1-relator presentation $\langle x_1,\ldots, x_m ,y_1,\ldots, y_n | R\rangle$, such that $\langle x_1,\ldots, x_m \rangle =A \leq G, \langle y_1, \ldots, y_n \rangle=B \leq G$ (for this, one has to use the strong version of Grushko that any 1-relator presentation is Nielsen equivalent to one of this type). Suppose that $R$ is cyclically reduced, and involves a generator $x_1 \in F_m$. Then $\langle x_2,\ldots , x_m, y_1, \ldots, y_n\rangle$ generates a free subgroup of $G$ by the Freiheitssatz. But this implies that $B = \langle y_, \ldots, y_n\rangle$ is free. Moreover, if $R$ involves one of the generators $y_i$, then one sees that $A=\langle x_1,\ldots,x_m\rangle$ is also free, and therefore $G=A\ast B$ is free, a contradiction. So $R \in F_m$, as required.