It doesn't hurt to expand. (And its probably the only way as well.) \begin{align*}(1+\sqrt2)^5&=1+\binom51\sqrt2+\binom522+\binom532\sqrt2+\binom544+4\sqrt2\\ &=1+5\sqrt2+20+20\sqrt2+20+4\sqrt2\\ &=41+29\sqrt2\end{align*}The answer is $41+29=\boxed{70}$.

We can use the recursion. $(a+b\sqrt 2)(1+\sqrt 2)=a+2b+(a+b)\sqrt 2$. $$(1,1)\to (3,2)\to(7,5)\to(17,12)\to(41,29)$$$41+29=70$