Are there integral solutions for $(2a-1)(2^{(b+c)}-3^c )=2^b-1$? Can anyone prove this assertion?  Or at least suggest a method of attack?  It has come up in my research.
There do not exist $a,b$ and $c$ such that$$
 (2a-1)(2^{(b+c)}-3^c )=2^b-1
$$where $a>0,b>1,c>1$ and $a,b,c ∈ Z$
This question came up as I was comparing 2 types of sums: $S1=x+(3/2)x+(3/2)^2x+...+(3/2)^cx, S2=y+2y+2^2y+...+2^by$ to see if they could ever equal one another, given specific constraints on the relationship between x and y - specifically that $y=2x−1$ and $x=2^c(2a−1) $
 A: Let me rewrite the letters for your variables due to my long-time practice.
I usually write $N$ for $c$ , $S$ for $c+b$ such that $S = \lceil N \cdot \log_2(3) \rceil$ and $B$ for $b$.
Also, to simplify I write $k$ for $2a-1$, not forgetting $k$ must be odd.                
I refer also to a formula of G. Rhin, cited in J. Simons [Si,07], for a lower bound of $S \log 2 - N \log 3$ depending on $N$.             
So we start with your formula, simply rewritten in notation, and then adapting for application of Rhin's inequality:
$$ \begin{array} {rl} k(2^S-3^N)&=2^B - 1 \\
 k( (2^S/3^N)-1)&=(2^B - 1)/3^N \\
 2^S/3^N &= 1 + (2^B - 1)/k/3^N 
\end{array} \tag 1$$
Logarithmizing and using the Mercatorseries on the rhs gives us
$$\small {S \ln2 - N \ln3  =  (2^B - 1)/k/3^N -1/2((2^B - 1)/k/3^N)^2+ 1/3((2^B - 1)/k/3^N)^3 \pm \ldots }$$
Cancelling of trailing small terms on the rhs gives now an in-equality
$$ S \ln2 - N \ln3  \lt  (2^B - 1)/k/3^N \tag 2 $$
By G. Rhin we have due to J. Simons (formula slightly rewritten to be better memorizable):
$$\frac1{457}\frac1{N^{13.3}} \lt S \ln2 - N \ln3 \tag 3
$$
So we can conclude
$$\begin{array} {rl}  \frac1{457}\frac1{N^{13.3}} &\lt S \ln2 - N \ln3  &\lt  (2^B - 1)/k/3^N \\
\frac1{457}\frac1{N^{13.3}} &\lt  (2^B - 1)/k/3^N 
\end{array} \tag 4  $$
Now taking logarithms again gives
$$\small \begin{array} {rl}  -6.13 - 13.3 \ln N &\lt \ln (2^B - 1) - \ln k - N \ln 3 \\ 
 N \ln 3 - 13.3 \ln N &\lt 6.13 + B \ln 2 -(1/2^B + 1/2/4^B + ..) - \ln k  \\
 N \ln 3 - B \ln 2 - 13.3 \ln N &\lt 6.13  -(1/2^B + 1/2/4^B + ..) - \ln k  \\
\end{array} \tag 5$$
Here obviously the lhs is increasing with increasing $N$ while the rhs stays roughly constant (or even decreases if you increase $k$) so we can solve for the equality-condition given $N$ and $\small{B=\lceil N \cdot (\log_2 (3) -1)\rceil }$ and some assumed $k$. Say $k=3$ then at $\small {N=95.05}$ the lhs grows over the rhs.        
So for $N \gt 95$ there is no solution.
The cases $N \le 95$ can be done one by one finding that no other solution exists for $N>2$       
Now check for other odd $k>3$.     
So you are done.

Remark: I think this is easier than the Simon's exposition because I do not need to refer to the theory of continued fractions here.

[Si,07]  John L Simons: On the (non)-existence of m-cycles for generalized Syracuse sequences
$ \qquad \qquad $(2007) (online update of earlier article)
[Rh,87]  G. Rhin: Aproximants de Padé et mesures d’irrationalité.
$ \qquad \qquad $Progress in Mathematics 71, (1987), pp. 155-164
$ \qquad \qquad $(Reference supplied by [Si,07])
