On the integer solutions of some exponential equations I am trying to find the non-negative integer solutions of the following exponential equations:
$2^a=3^b+1\quad\color{blue}{(1)}$
$3^a=2^b+1\quad\color{blue}{(2)}$
I have found a way to obtain the non-negative integer solutions of these two exponential equations, but I think there exists a most refined and shorter way.
Could you help me to get it?

I am going to solve the equation $\;(1)\;\;2^a=3^b+1\;$.
If $\;b=0\;$ then $\;a=1\;.$
If $\;b\ge 1\;$ then $\;a\;$ is even.
I prove it by contradiction. If $\;a\;$ were odd, it would exist $\;\alpha\in\mathbb{N}_{0}=\mathbb{N}\cup\{0\}$ such that $\;a=2\alpha+1.$ Consequently, $\;2\cdot 4^\alpha=3^b+1\;,$ hence $\;2\left(3+1\right)^\alpha=3^b+1\;.$ So there exists $\beta\in\mathbb{N}_{0}\;$ such that $\;2\left(3\beta+1\right)=3^b+1\;,$ that is $\;1=3\left(3^{b-1}-2\beta\right),$ but the last equality contradicts the fact that $\;1\;$ is not a multiple of $\;3\;.$ Therefore $\;a\;$ is even otherwise it would lead to a contradiction.
Since $\;a\;$ is even, there exists $\;\alpha\in\mathbb{N}\;$ such that $\;a=2\alpha\;,\;$ so the equation $\;(1)\;$ turns into the following one:
$4^\alpha-1=3^b\;,$
$\left(2^\alpha+1\right)\left(2^\alpha-1\right)=3^b\;.$
Hence,
$\begin{cases}
  2^\alpha+1 &= 3^h\\
  2^\alpha-1 &= 3^{b-h}
\end{cases}\quad$
where $\;h\in\mathbb{N}\;$ and $\;h\le b\;.$
By subtracting the previous equalities, we get that
$2=3^{b-h}\left(3^{2h-b}-1\right)\;.$
Hence $\;h=b\;$ and $\;3^b-1=2\;$.
So $\;b=1\;,\;\alpha=1\;$ and $\;a=2\;.$
Consequently the non-negative integer solutions of the equation $(1)$ are $\;a=1\;,\;b=0\;$ and $\;a=2\;,\;b=1\;.$

Now I am going to solve the equation $\;(2)\;\;3^a=2^b+1\;$.
Since $\;a\in\mathbb{N}\;$ there exist $\;n\in\mathbb{N}_{0}\;$ and $\;a_1\in\mathbb{N}\;$ ($a_1$ odd) such that $\;a=2^n a_1\;.$
So the equation $(2)$ turns into the following one:
$\left(3^{2^n}\right)^{a_1}-1=2^b\;,$
$\left(3^{2^n}-1\right)\cdot\left[1+3^{2^n}+\left(3^{2^n}\right)^2+\ldots+\left(3^{2^n}\right)^{a_1-1}\right]=2^b.$
Since inside the square brackets there is a sum of a number odd $\;a_1\;$ of addends which are all odd, the sum inside the square brackets is odd, so the sum has to be equal to $\;1\;$ otherwise it could not be a factor of $\;2^b\;,\;$ consequently $\;a_1=1\;$ and $\;a=2^n\;.$
$3^{2^n}-1=2^b\;.$
If $\;n=0\;$, then $\;a=1\;$ and $\;b=1\;.$
If $\;n\ge1\;$ the last equality turns into the following one by decomposing the left-hand side into factors:
$\left(3^{2^{n-1}}+1\right)\left(3^{2^{n-2}}+1\right)\cdots\left(3^{2^1}+1\right)\left(3^{2^0}+1\right)\left(3^{2^0}-1\right)=2^b\;,$
Since $\;\left(3^{2^1}+1\right)=2\cdot 5\;$ and $\;5\;$ is not a factor of $\;2^b\;$, it follows that $\;n-1<1\;,\;$ hence $\;n=1\;,\;a=2\;$ and $\;b=3\;.$
Consequently the non-negative integer solutions of the equation $(2)$ are $\;a=1\;,\;b=1\;$ and $\;a=2\;,\;b=3\;.$

Could you help me to get a most refined and shorter solution to the two exponential equations?

Here's a condensed proof:
If $a$ and $b$ are integers such that $2^a=3^b+1$ then reducing mod $3$ shows that $a$ is even if $b>0$, say $a=2c$. Then
$$3^b=2^a-1=2^{2c}-1=(2^c-1)(2^c+1),$$
and so the two factors are both powers of $3$. They differ by $2$ so they are $1$ and $3$, so $b=c=1$, and $a=2$. If $b=0$ then clearly $a=1$.
Similarly, if $a$ and $b$ are integer such that $3^a=2^b+1$ then reducing mod $4$ shows that $a$ is even if $b>1$, say $a=2c$. Then
$$2^b=3^{2c}-1=(3^c-1)(3^c+1),$$
and so the two factors are both powers of $2$. They differ by $2$ so they are $2$ and $4$, so $c=1$ and $a=2$ and $b=3$. It remains to check for solutions with $b\leq0$. A quick check yields only $(a,b)=(1,1)$.
