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",.)

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.
