<s>[INST] Find $x$ and $y$ in $2^{x-y} + 1 = 2^x,$ where $x,y$ are integers I have no idea what to do now. 
Is there any way to find the integers $x$ and $y$ by factoring?
Thank you.
 [/INST] As an alternative (using factoring), note that $2^{x-y}=\frac{2^x}{2^y},$ so multiplying both sides by the non-zero number $2^y$ yields the equivalent equation $$2^x+2^y=2^x2^y,$$ which in turn yields $$0=2^x2^y-2^x-2^y\\0=2^x(2^y-1)-2^y\\0=2^x(2^y-1)-2^y+1-1\\0=2^x(2^y-1)-(2^y-1)-1\\0=(2^x-1)(2^y-1)-1\\1=(2^x-1)(2^y-1)$$
Note that we cannot have $x=0$ or $y=0$. (Why?) Also $x$ and $y$ must have the same sign, for if not, then $(2^x-1)(2^y-1)$ is negative. If $x,y$ are negative, then $(2^x-1)(2^y-1)$ is positive, but less than $1,$ so we must have $x,y$ positive.
Now, for any positive integers $x,y,$ we have $x,y\ge 1$ (since $x,y$ integers), so $2^x,2^y\ge 2,$ so $2^x-1,2^y-1\ge 1,$ and so $(2^x-1)(2^y-1)\ge 1.$ Equality holds in the last case precisely when it holds in the first case--that is, precisely when $x,y=1.$
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