<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] Here's a somewhat simpler proof using Pell equations
(as Eric Towers suggested might be possible).
If $x$ is odd, say $x=2k+1$, then $3z^2 = 2y^2 + 1$ where $z = 3^k$.
Hence $(2y)^2 - 6z^2 = -2$ and we have
$$
2y + 3^k \sqrt{6} = (2+\sqrt{6}) (5 + 2\sqrt{6})^m
$$
for some $m=0,1,2,\ldots$.  The first two solutions $m=0$, $m=1$ give
$z=1$, $z=9$ which recovers the known solutions with $x=1$ and $x=5$.
Suppose $k>2$.  Then $z = 3^k \equiv 0 \bmod 27$, and by computing powers of
$5 + 2 \sqrt{6}$ mod $27$, we find that $27 \mid z$ if and only if
$m \equiv 4 \bmod 9$.  But then $z$ is always a multiple of the $m=4$ solution
(in general if $2m+1 \mid 2m'+1$ then the $m$-th $z$ divides the $m'$-th one),
and the $m=4$ solution has $z = 8721 = 3^3 17 \, 19$.  So $z$ can never be
a power of $3$ once $k>2$.
If $x$ is even, say $x=2k$, then $2y^2 = 3^{2k} - 1 = (3^k-1) (3^k+1)$.
Hence $3^k-1$ is either a square or twice a square, and the former is
impossible (no square is congruent to $-1 \bmod 3$).  So $3^k-1 = 2{y'}^2$
for some integer $y'$, giving a smaller solution $(k,y')$ to $3^x = 2y^2 + 1$.
Continuing in this fashion eventually yields a solution with $x$ odd. 
But that solution cannot be $(x,y) = (5,11)$, because
$(3^{10} - 1) / 2 = 2^2 11^2 61$ is not a square.  This leaves
the $(x,y) = (1,1)$ solution, and indeed $(x,y) = (2,2)$ also works
as we know $-$ but $x=4$ does not because $(3^4 - 1)/2 = 40$
is not a square either.  This completes the proof that the three solutions
with $x=1,2,5$ are the only ones. 
</s>