I'd like to have at hand some simple cases (while outstanding famous examples are provided b y literature and Internet).
Let me render this equation in $\LaTeX$: $$2^x-3^y=1$$ That's better. We consider here non-negative integers $x\ y$ only. It's easy to guess the whole set of solutions:
We will see that there are no more solutions but the above two. First of all let's observe that for any value of $y$ (or respectively for $x$) there is at the most value of $x$ (respectively of $y$) such that the pair $(x\ y)$ is a solution, i.e. such that $2^x-3^y=1$. In particular, there are no solutions when $y=2$.
Let's consider exponents $x\mod 6$. As you will see in a moment three classes $\mod 6$ will be treated as one case, then the remaining three classes will form separate three cases, for a total of four. Each case will be decided by the respective congruence at the basic (not exponent) level:
Case 1. $\ \forall_{y\ge 1}\ \ 2^{2\cdot a+1} - 3^y\equiv 2\mod 3\ \ $ hence $2^{2\cdot a+1} - 3^y\ne 1$;
Case 2. $\ \forall_{y\ge 2}\ \ 2^{6\cdot a+2} - 3^y\equiv 4\mod 9\ \ $ hence $2^{6\cdot a+2} - 3^y\ne 1$;
Case 3. $\ \forall_{y\ge 2}\ \ 2^{6\cdot a+4} - 3^y\equiv 7\mod 9\ \ $ hence $2^{6\cdot a+4} - 3^y\ne 1$;
Case 4. $\ \forall_{y\ge 2}\ \ 2^{6\cdot a} - 3^y\equiv 1 - 3^y\mod 9\ \ $ hence $2^{6\cdot a} - 3^y\ne 1$;
Thus there are no solutions besides the two entioned above:
THEOREM 0 An ordered pair $(x\ y)$ of non-negative integers satisfies equality $2^x-3^y=1\ \Leftrightarrow\ \ (x\ y)=(1\ 0)$ or $(2\ 1)$.
REMARK 0 In Case 4 above we could consider $2^{3\cdot b}$ instead of $2^{6\cdot a}$. This would cover two residue classes $\mod 6$ (but it'd be a confusing overkill, hey!). The whole proof above seems to be somewhat ad hoc. Nothing terrible about except that I remember getting a nicer one in the past (years ago). I am sure that working $\mod 7$ at the basic level would be preferable, I might try it later.
Let me render this equation in $\LaTeX$: $$2^x-5^y=1$$ That's better. We consider here non-negative integers $x\ y$ only. It's easy to guess the whole set of solutions:
and there are no other solutions. Of course, this $(x\ y) = (1\ 0)$ is the only solution for $y=0$. Now assume that $y>1$. There exists non-negative integers $(d\ r)$ such that $x=4\cdot d + r$, where $0\le r < 4$. Consider the following computation $\mod 5$:
$$2^x-5^y\equiv 2^{4\cdot d+r}=16^d\cdot 2^r\equiv 2^r\mod 5$$
Thus for $y>0$ we get:
$$2^x-5^y\equiv 1\mod 5\quad\Leftrightarrow\quad r=0$$
i.e. when the equality holds,
$$2^x = 16^d\equiv 1 \mod 3$$
hence
$$1 = 2^x-5^y\equiv 1-5^y\mod 3$$
i.e. $5^y\equiv 0\mod 3$, a contradiction. This proves
THEOREM 1 An ordered pair $(x\ y)$ of non-negative integers satisfies equality $2^x-5^y=1\ \Leftrightarrow\ \ (x\ y)=(1\ 0)$.