$$\begin{align}
\left(9\over7\right)^3={729\over343}\gt2
&\implies\left(9\over7\right)^{15}\gt2^5\gt7\\
&\implies9^{15}\gt7^{16}\\
&\implies9^{90}\gt7^{96}\\
&\implies9^{91}\gt7^{94}
\end{align}$$
Another proof, using the general inequality $\ln(1-x)\lt-x$ for $0\lt x\lt1$ and the numerical inequality $7\lt2^3\lt e^3$:
$$\ln(7^{94}/9^{91})=3\ln7+91\ln\left(1-{2\over9}\right)\lt3\cdot3-90\cdot{2\over9}=9-20\lt0$$
A third proof, presented in easily checkable, but almost completely unmotivated form:
$$\begin{align}
2^{47}7^{94}
&=98^{47}\\
&\lt100^{47}\\
&=10^3\cdot10^3\cdot10^{88}\\
&\lt2^{10}\cdot2^{10}\cdot10000^{22}\\
&\lt20000^{22}\\
&\lt160^{44}\\
&\lt36(162)^{45}\\
&=2^{47}9^{91}
\end{align}$$
And one more proof, this one based on the fact that $2^{10}=1024\gt1000=10^3$, which implies $\log2\gt0.3$ (where "log" here means log base $10$):
$$94\log7=47\log49\lt47(\log100-\log2)\lt47\cdot1.7=79.9$$
whereas
$$91\log9\gt91\log8=273\log2\gt273\cdot0.3=81.9$$
Full disclosure: I used a calculator for $47\cdot1.7=79.9$. But everything else I did by hand.
Added 5/25/15: At another question, proofs are given of the inequality $7^{19}\lt9^{17}$. (See in particular joriki's answer there.) It follows that
$$7^{94}\lt7^{95}\lt9^{85}\lt9^{91}$$