\[\begin{array}{l} {\text{Пусть }}k > 1,{\text{ }}m > 1 \hfill \\ {7^k} = {3^m} + 4 \Leftrightarrow 7\left( {{7^{k - 1}} - 1} \right) = 3\left( {{3^{m - 1}} - 1} \right) \hfill \\ {\operatorname{ord} _7}3 = 6 \Rightarrow {2^3} \cdot 7 \cdot 13 = {3^6} - 1|{3^{m - 1}} - 1 \hfill \\ {\operatorname{ord} _{13}}7 = 12 \Rightarrow {3^2}|{7^{12}} - 1|{7^{k - 1}} - 1.{\text{ Противоречие}}{\text{.}} \hfill \\ \left( {k;m} \right) = \left( {1;1} \right). \hfill \\ \end{array}\]

\[\left( {k;m} \right) = \left( {1;1} \right)\]