Trivial with Hensel's Lemma! The number of factors $2$ of $3^{x}-1$ is $f(x)+1$, where $f(x)$ is the number of factors $2$ of $x$. So, $x \leq f( x )+1$, that implies x = 0 or 1

We have only $x\le ord_{2}(3^{x}-1)$. If x odd, then $x=1$. If $x=2^{z}m, z\ge 1$, were m - odd, then $ord_{2}(3^{x}-1)=z+2$, therefore $2^{z}m\le z+2$. It give $m=1, z=1 \ or \ z=2$. Solutions are $x=0,y=0$, $x=1,y=1$, $x=2,y=2$, $x=4,y=5$.

Could some one explain Hensel's formula? What does he mean by number of factors 2?

My solution: We easy get $x=0,y=0,x=1,y=1,x=2,y=2,x=4,y-5$, and for $x=3, x=6$ we do not have solutions. Now suppose$x\ge 7$. We have of course that $x$ is even(easily get mod8), so $x=2k$, we have $(3^k-1)(3^k+1)=2^{2k}y$, but $gcm((3^k-1),(3^k+1)=2)$ So, let's see what happens when k is even. We get $k=2^ut,gcm(t,2)=1.$$3^k+1=2(mod4)$, so denote $v_2(w)$ the exponent of $2$ in $w$, we have $v_2(3^k-1)\ge 2k-1$, but $3^{2^ut}-1=(3^t-1)(3^t+1)(3^{2t}+1)...(3^{2^{u-1}t}+1)$, but we easy get $3^{2x}+1=2(mod4)$ so the inequality can be written ${u+v_2(3^t+1)\ge 2^{u+1}t-1}$, obvios false, because $3^t+1\le 4^t$, and $v_2(4^t)=2t$. So, we are done when $k$ is even. Suppose now that $k$ is odd, $3^k-1=2(mod4)$, so we have${v_2(3^k+1)\ge 2k-1}$, which is false(analogous with what i've done before)

See here (from 2005 Romanian TST) : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=198299&sid=8e00e8b252767dc670e5c9bd7a604079#p198299

Agr_94_Math wrote: Could some one explain Hensel's formula? What does he mean by number of factors 2? This is actually well-known as Lifting The Exponent Lemma

Can you also solve it by dividing by 2^x and taking away 2^-x and show x has to be small for there to be no fractions in the binomial expansion?

The equation $3^x=2^xy+1$ implies that $3^x\equiv 1$(mod $2^x$) and by Euler-Fermat $x\equiv 0$ (mod$2^{x-1}$) which means that $2^{x-1}\le x$, so $x\le2$. Substituting, we get that two only possible solutions are $(1,1)$ and $(2,2)$.

N.T.TUAN wrote: Find all natural solutions of $3^{x}= 2^{x}y+1.$ We claim that the only solutions are $\boxed{(1,1)},\boxed{(2,2)},\boxed{(4,5)}$.We now prove that these are the only. If $x=1$ then we get $(x,y)=(1,1)$ is a solution otherwise for $x\geq 2$ we get $x$ is even by looking modulo $4$.Now $$\nu_2(3^x-1)=\nu_2(3+1)+\nu_2(3-1)+\nu_2(x)-1=\nu_2(x)+2$$.Hence $\nu_2(x)+2\geq x \Longleftrightarrow x\leq 2+\lceil \log_2x \rceil \implies \lceil \log_2x \rceil\geq x-3 \implies x\geq 2^{x-3}$ which clearly fails for $x\geq 6$.Thus $x<6$.Now checking the permissible range we get the claimed solutions.$\blacksquare$

We have trivial solutions $(x,y)=(0,0);(1,1);(2,2)$. We need to find another solution. See that $3^x \equiv 1 \pmod {16}$ if $3^x \equiv 0 \pmod {4}$, so we can write $x=4k$, for some $k$. Our equation becomes $3^{4k}-1=2^{4k}y$ or $(3^{2k}-1)(3^{2k}+1)=2^{4k}y$. It is easy to see that $y$ is odd or even. So we have two cases $I$ case is $y \equiv 0 \pmod {2}$: Let's define $z$ and $d$ such $zd=y,$ so $3^{2k-1}=4^{2k}z$ and $3^{2k}+1=d$. This has obviously no solution. $II$ case is $y \equiv 1 \pmod {2}$: Again, $3^{2k}=4^{2k-1}2z$ and $3^{2k}+1=2d.$ This has solutions when $k<3$ and that is for $k=1 \implies z=1, d=5$ and finally $x=4, y=5$. So, all solutions are $(x,y)=(0,0);(1,1);(2,2);(4,5)$