<s>[INST] Find all natural numbers $x,y$ such that $3^x=2y^2+1$. Find all natural numbers $x,y$ such that 
$$3^x=2y^2+1$$
solutions are $(1,1)$, $(2,2)$, $(5,11)$. I found that parity of both is same and If $x$ Is odd it is of the form $4k+1$. 
 [/INST] There is a "standard" way to attack this using Thue's theorem, by writing the left hand side as $z^3$, $3 z^3$, or $9 z^3$, depending on the assumed residue $x \pmod 3$.  I can't imagine doing a search up to any of the effective versions of Thue's bound on the size of the solutions during an Olympiad.
Some of the comments are close to suggesting that we attack this as a Pell equation, so I have started that below.  I don't know how to finish either case.  Perhaps one of the commenters would be willing to edit in a completion to either case below.
If $x$ is even, $x = 2k$, this is the Pell equation $(3^k)^2 - 2 y^2 = 1$.  The usual technique for finding the minimal solution, constructing convergents of the continued fraction for $\sqrt{2}$, gives the integer solutions $$\begin{align}
    3^k &= \frac{1}{2} \left( (3-2\sqrt{2})^n + (3+2\sqrt{2})^n \right) \\
    y   &= \frac{-1}{2\sqrt{2}} \left( (3-2\sqrt{2})^n - (3+2\sqrt{2})^n \right),
\end{align}$$ for any integer $n$ (and not yet applying the restriction that $k$ be a positive integer).  Note that $y>0$ requires $n \geq 0$.  In fact, $n=0$ gives $3^k = 1$, so $k$ and $x$ are both zero (and $x = 0$ is not allowed).  Hence, $n > 0$.
If $x$ is odd, $x = 2k+1$, we multiply through by $3$ to get $3^2 (3^k)^2 - 6 y^2 = 3$, another Pell equation, whose integer solutions are $$\begin{align}
    3 \cdot 3^k &= \frac{1}{2}\left( (3+\sqrt{6}) (5-2\sqrt{6})^n + (3 - \sqrt{6})(5+2\sqrt{6})^n \right) \\
    y           &= \frac{-1}{4} \left( (2+\sqrt{6})(5-2\sqrt{6})^n + (2 - \sqrt{6})(5+2\sqrt{6})^n\right),
\end{align}$$ for any integer $n$ (and not yet applying the restriction that $k$ be a positive integer).  Note that $y>0$ requires $n>0$.
It's not clear to me that the $k \in \Bbb{N}$ restriction is straightforward to apply to either of these.  But I would be happy for someone(s) to edit in a completion to either of the above.
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