<s>[INST] 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 $4\mid u$.
 [/INST] 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:
$$ \begin{array} {rlll}
   -9^t &= -1 &\not \equiv 1 \pmod 8 &\qquad \text{for odd $v=1+2t$} \\
    -3\cdot 9^t &= -3& \not \equiv 1 \pmod 8 &\qquad \text{for even $v=2t$} \\
\end{array}$$    
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