Logarithmic equation nonstandard.

Question

$$\log_{7 }(6 ^x+1) =\log_{3 }(4^x-1) $$ How can I prove that $x=1$ is the only solution without derivatives? If anyone can give me a hint or some point to start from I would appreciate. I found a solution.We know that $4^x-1>0 \Leftrightarrow x>0$ $$\frac{\log(6 ^x+1)}{\log7}=\frac{\log(4^x-1)}{\log3}$$ $$\Leftrightarrow\log3 \cdot \log(6^x+1)=\log7 \cdot \log(4^x-1)$$ We set $f(x)=\log3 \cdot \log(6^x+1)-\log7 \cdot \log(4^x-1)$and take $x_{1}>x_{2}$. I evaluate $$f(x_{1})-f(x_{2})=\log3 \cdot \left [ \log(6^{x_{1}}+1)-\log(6^{x_{2}}+1)\right ]-\log7 \cdot \left [ \log(4^{x_{1}}-1)-\log(4^{x_{2}}-1)\right ]$$ $$f(x_{1})-f(x_{2})=\log3 \cdot \left [ \log{\frac{(6^{x_{1}}+1)}{(6^{x_{2}}+1)}}\right ]-\log7 \cdot \left [ \log{\frac{(4^{x_{1}}-1)}{(4^{x_{2}}-1)}}\right ]$$ $$f(x_{1})-f(x_{2})=\log3 \cdot \left [ \log{\frac{6^{x_{1}}(1+\frac{1}{6^{x_{1}}})}{6^{x_{2}}(1+\frac{1}{6^{x_{2}}})}}\right ]-\log7 \cdot \left [ \log{\frac{4^{x_{1}}(1-\frac{1}{4^{x_{1}}})}{4^{x_{2}}(1-\frac{1}{4^{x_{2}}})}}\right ]$$ $$f(x_{1})-f(x_{2})=$$$$\log3 \cdot \log6^{x_{1}-x_{2}}+\log3 \cdot \log{\frac{(1+\frac{1}{6^{x_{1}}})}{(1+\frac{1}{6^{x_{2}}})}}-\log7 \cdot \log4^{x_{1}-x_{2}}-\log7 \cdot \log{\frac{(1-\frac{1}{4^{x_{1}}})}{(1-\frac{1}{4^{x_{2}}})}}$$ $$(x_{1}-x_{2}) \cdot(\log3 \cdot \log6- \log7 \cdot \log4)+\log3 \cdot \log{\frac{(1+\frac{1}{6^{x_{1}}})}{(1+\frac{1}{6^{x_{2}}})}}+log7 \cdot \log{\frac{(1-\frac{1}{4^{x_{2}}})}{(1-\frac{1}{4^{x_{1}}})}}<0$$ So $f(x)$ is strictly decreasing and it is injective that means the equation has at most one solution that is $x=1$.