How to find all nonnegative integers $x, y, z$ and $w$ such that $2^x3^y-5^z7^w=1$ 
Find all nonnegative integers $x, y, z$ and $w$ such that $2^x3^y-5^z7^w=1.$

I think they are $(x,y,z,w)=(1,0,0,0),(1,1,1,0),(3,0,0,1),(2,2,1,1)$, but I couldn't prove its sufficiency (or there may be other solutions).
Does anybody have a good (or elegant) solution to this?
(Be careful that $x,y,z,w$ are "nonnegative integers",.)
 A: One possible solution is as follows: you're looking for two consecutive numbers where all prime factors are 7 or less.  The list of numbers $n$ such that $n$ and $n+1$ have all their prime factors 7 or less is finite, and can be found at OEIS; it's
1, 2, 3, 4, 5, 6, 7, 8, 9, 14, 15, 20, 24, 27, 35, 48, 49, 63, 80, 125, 224, 2400, 4374.
The fact that this sequence is finite is Størmer's theorem, which also provides an algorithm for finding the list.
Furthermore, you want a number in this sequence, $n = 5^z 7^w$, with prime factors 5 and 7, such that $n + 1$ only has prime factors 2 and 3.  The numbers in this sequence with prime factors 5 and 7 only are (by inspection):
1, 5, 7, 35, 49, 125
Adding 1 to each of these, you can see that 2, 6, 8, 36 have prime factors 2 and 3 only, while 50 and 126 do not.  So the only solutions are $n = 1, 5, 7, 35$ which correspond to your four solutions.
A: Imagine all are bigger than 0, also, imagine x is bigger than 2. then modulo 8 $((-3)^z)((-1)^w)+1=0$ Then, given that z is odd because of mod 3, we get that in either case, be it w even or odd there is a contradiction, so x is either 1 or 2.
Be it 1, modulo 7 we find there should be this modularity pairs (x,y) = (3m, 6k) (3m+1,6k+4) (3m+2,6k+2), and modulo 5 we find they are (4u,4t) (4u+1,4t+1) (4u+2,4t+2) (4u+3,4t+3). So we get a contradiction brcause then 6k+4=4t+1. Imagine then x=2. Then, if both z and w are 2 or more, modulo 25 we get y=60Q+14 and modulo 49 we get y=42m+21 wich is a contradiction, so both z and w can not be both 2 or more.
We now have bounds so the rest is casework
