My solution from the contest. A bit computational, but it works. So, for $n=2$ one may check that $a=\frac{1}{2}-\sqrt{2}$ works. Henceforth suppose $n\ge 3$. $\textbf{Lemma.}$ If $p,q,r,s$ are rational numbers such that $p+q\sqrt{2}=r+s\sqrt{2}$, then $p=r$ and $q=s$. $\textbf{Proof.}$ Suppose the contrary. Then $\sqrt{2}=\frac{p-r}{s-q}$ exists and is rational, a contradiction. $\blacksquare$ Denote $a+\sqrt{2}=l$ and $a^n+\sqrt{2}=k$, where $k$ and $l$ are rational numbers. Consequently we have $(k-\sqrt{2})^n+\sqrt{2}=l$, or equivalently written $$k^n-{{n}\choose{1}} k^{n-1}\sqrt{2}+{{n}\choose{2}}k^{n-2}(\sqrt{2})^2+\dots+(-\sqrt{2})^n+\sqrt{2}=l.$$This suggests us to consider 2 cases, namely $n=2u$ and $n=2u+1$. If $n=2u$, $u>1$, one can easily check that we must have, by the lemma, $${{2u}\choose{1}}k^{2u-1}+{{2u}\choose{3}}k^{2u-3}\cdot 2+\dots +{{2u}\choose{2u-1}}k2^{u-1}=1.$$Let $k=\frac{p}{q}$ where $gcd(p;q)=1$, $q>0$ and p is an integer. Then by the rational root theorem we get $p=1$ and $q|2u$ (note that we must have $p>0$ in order the equation to have both members positive). However, we also get that $2u\cdot q^{2u-2}\cdot 2^{u-1}<q^{2u-1}$, which combined with $q\leq 2u$ gives $2^u<2$, false. If $n=2u+1$, $u>0$, again the lemma tells us that $${{n}\choose{1}}k^{n-1}+{{n}\choose{3}}k^{n-3}2^1+\dots +{{n}\choose{n}} 2^u=1.$$All terms being positive ($n$ is odd), we shall have ${{n}\choose{n}}\cdot 2^u<1$, vacuously false. Hence the only solution is $n=2$, exemplified at the beginning of the solution.

Very good problem, roughly same soln as @above , in @above's solution $k$ and $l$ seem swapped.