Number of solutions of $2^u=3^l+1$ I want to prove that the unique solution for this equation with $u,l\ge1$ are $u=2$ and $l=1$. I only can prove there doesn't exist a solution if $3\mid u$ or $4\mid u$.
 A: We want to show there is no integer solution with $\ell >1$. 
The first claim: $$3^\ell || 2^{2*3^{\ell -1} }-1,$$ where $p^x || y$ means $p^x|y$ and $p^{x+1} \nmid y$.   
The proof can be done in several ways probably, and induction should work, by using $$x^3-1 = (x-1)(x^2+x + 1).$$
Then if $2^u=3^\ell + 1$, we must have $u\mid 2*3^{\ell -1}$, and $u$ is of form $3^x$ ( which is in fact not ok since $3 \nmid 2^{3^x} - 1$ ) or $2*3^x$, and by the claim, we must have $u=2*3^{\ell -1}$. 
But when $\ell > 1$, $$2^u = 2^{2*3^{\ell -1}} > 3^\ell + 1$$
A: The powers of $3$ mod $8$ are $1$ and $3$.  Thus $3^l+1\in\{2,4\}$ mod $8$.  It is therefore not possible to have $2^u=3^l+1$ if $u\ge3$.  This leaves only $u=0,1$, and $2$ to be considered.  Only the pairs $(u,l)=(1,0)$ and $(2,1)$ work.
My great thanks to Gottfried Helms for pointing out, in comments, a significant simplification to the initial answer here.
A: I guess I should write this simpler solution. We must have $2\mid u$ given that $3 \mid 2^u -1$. Write $u=2x$, then $3^\ell=2^u-1 = (2^x+1)(2^x-1)$, given that $gcd(2^x+1, 2^x -1)=gcd(2^x+1, 2)=1$, we must have $2^x-1=1$, as otherwise either $2^x+1$ or $2^x-1$ is not divisible by $3$. 
A: We have one solution $2^2 - 3^1 =1 $ which is also $2^2 -3^1 =3^2-2^3 $
Looking for more solutions, with $u,v \ge 1$
 $$2^{2+u} -3^{1+v} =3^2-2^3  $$
Looking at this modulo $8$ 
 $$2^{2+u} -3^{1+v} \overset{?}\equiv 3^2-2^3  \pmod 8 \\
    0 -3^{1+v} \overset{?}\equiv 3^2- 0  \pmod 8 \\
    -3^{1+v} \overset{?}\equiv 1  \pmod 8 $$
For odd and even $v$ separated shows, there is no solution:
$$ \begin{array} {rlll}
   -9^t &= -1 &\not \equiv 1 \pmod 8 &\qquad \text{for odd $v=1+2t$} \\
    -3\cdot 9^t &= -3& \not \equiv 1 \pmod 8 &\qquad \text{for even $v=2t$} \\
\end{array}$$    
