No calculator, please!

Expanding, we have that $(\sqrt{2} + 1)^{3} = (\sqrt{2})^{3} + 3*(\sqrt{2})^{2} + 3*\sqrt{2} + 1 = 7 + 5\sqrt{2},$ and so

$(\sqrt{2} + 1)^{6} = (7 + 5\sqrt{2})^{2} = 49 + 25*2 + 2*5*7*\sqrt{2} = 99 + 70\sqrt{2}.$

Now $2 \gt 1.96 = (1.4)^{2}$ , so $\sqrt{2} \gt 1.4$ . This in turn implies that

$99 + 70\sqrt{2} \gt 99 + 70*1.4 = 99 + 98 = 197.$

Next, note that $(70\sqrt{2})^{2} = 4900*2 = 9800 \lt 9801 = 99^{2},$ and so $70\sqrt{2} \lt 99$ Thus

$99 + 70\sqrt{2} \lt 99 + 99 = 198.$

Combining these results, we see that $197 \lt 99 + 70\sqrt{2} \lt 198$ , and so $\lfloor (\sqrt{2} + 1)^{6} \rfloor = \boxed{197}.$

Note that $\left( \sqrt{2}+1 \right)^6 + \left( \sqrt{2}-1 \right)^6 = 2 \left( \left( \sqrt{2} \right)^6 + 15 \cdot \left( \sqrt{2} \right)^4 + 15 \cdot \left( \sqrt{2} \right)^2 + 1 \right) = 198$ , and $0 < \left( \sqrt{2}-1 \right)^6 < 1$ is obvious. Thus $197 < \left( \sqrt{2}+1 \right)^6 < 198$ , so $\left\lfloor \left( \sqrt{2}+1 \right)^6 \right\rfloor = \boxed{197}$ .

Why does the first equality work? It's just breaking open each of them by binomial theorem. The odd exponents include a negative sign in $(\sqrt{2}-1)^6$ , canceling the ones in $(\sqrt{2}+1)^6$ , so all that's left are even exponents which give integral (and thus easily computable) values. Also, it's convenient that $\sqrt{2}-1 \in (0,1)$ , so that raising it to any positive power still gives a small number, which gets canceled by the floor.

$(\sqrt2 + 1)^6 = ((\sqrt2 + 1)^2)^3 = (3 + 2 \sqrt2)^3 = 27 + 3 (3)^2 (2 \sqrt2) +3 (3) (2 \sqrt2)^2 + (2 \sqrt2)^3$

$\implies 27 + 54 \sqrt2 + 72 + 16 \sqrt2 = 99 + 70 \sqrt2 \approx 99 + 7 \times 14.14 = 99 + 98.98 = 197.98$

$\implies \lfloor 197.98 \rfloor = 197$

Answer: $\boxed{197}$