Beauty is critical!

Note that:

$\dfrac {3^{201}+2^{201}}{3^{197}+2^{197}} = \dfrac {81(3^{197})+16(2^{197})}{3^{197}+2^{197}} < \dfrac {81(3^{197})+81(2^{197})}{3^{197}+2^{197}} = 81$

And:

$\begin{aligned} \frac {3^{201}+2^{201}}{3^{197}+2^{197}} & = \frac {{\color{#3D99F6}81(3^{197})}+16(2^{197})}{3^{197}+2^{197}} \\ & = \frac {{\color{#3D99F6}80(3^{197}) + 3^{197}}+16(2^{197})}{3^{197}+2^{197}} \\ & = \frac {80(3^{197}) + {\color{#3D99F6}(2+1)^{197}}+16(2^{197})}{3^{197}+2^{197}} & \small \color{#3D99F6} \text{By binomial series expansion} \\ & = \frac {80(3^{197}) + {\color{#3D99F6}2^{197} + 197(2^{196})+ \cdots}+16(2^{197})}{3^{197}+2^{197}} \\ & = \frac {80(3^{197}) + {\color{#3D99F6}2^{197} + 98.5(2^{197})+ \cdots}+16(2^{197})}{3^{197}+2^{197}} \\ & > \frac {80(3^{197}) + 80(2^{197})}{3^{197}+2^{197}} = 80 \end{aligned}$

Therefore, $80 < \dfrac {3^{201}+2^{201}}{3^{197}+2^{197}} < 81$ $\implies \left \lfloor \dfrac {3^{201}+2^{201}}{3^{197}+2^{197}} \right \rfloor = \boxed{80}$ .

$\large \color{#3D99F6} \dfrac{3^{201} + 2^{201}}{3^{197} + 2^{197}} \approx \color{#D61F06} 3^{(201 - 197)} - (3 - 2) \color{#20A900} \implies 3^4 - 1 = \boxed{ 80 }$