Solutions to $2^a3^b+1=2^c+3^d$ 
Find all $a,b,c,d$ positive integer such that:
$2^a3^b+1=2^c+3^d$

My progress:
One solution satisfying is $$\boxed{a=1,b=1,c=2,d=1} $$
We first take $\mod 3$ which gives $$ L.H.S\equiv 1\mod 3,~~R.H.S\equiv 2^c\mod 3$$
Hence we get $c$ even. So let $c=2k.$
We get $$2^a3^b-3^d=2^{2k}-1=(2^k-1)(2^k+1). $$
Now note that $2^k-1,2^k+1$ are relatively prime. Because, if not then let $d$ be the common divisor.
Then $$d|2^k-1,~~d|2^k+1\implies d|(2^k+1)-(2^k-1)=2\implies d|2^k\implies d|2^k+1-2^k=1.$$
Now there are two cases. So using the fact that $2^k-1,2^k+1$ are relatively prime and then for odd k $3|2^k+1$ and for even $k$ $3|2^k-1$
Case 1: When $d<b$ then $$2^a3^b-3^d=3^d(2^a3^{b-d}-1)=(2^k-1)(2^k+1)$$

*

*$K$ is odd $$\implies v_3(2^k+1)=d,~~3\nmid 2^k-1. $$

*$K$ is even $$\implies v_3(2^k-1)=d,~~3\nmid 2^k+1. $$
Case 2: When $d>b$ then $$2^a3^b-3^d=3^b(2^a -3^{d-b})=(2^k-1)(2^k+1)$$

*

*$K$ is odd $$\implies v_3(2^k+1)=b,~~3\nmid 2^k-1. $$

*$K$ is even $$\implies v_3(2^k-1)=b,~~~~3\nmid 2^k+1. $$

P.S. This is my 100th post in MSE. The other solutions are there in the chat. Any elementary method?

This is not an answer, it is a comment for discussion:
We can construct similar equation as follows:
$$\begin{cases}3^m-2^n=1\\3^r-2^s=1\end{cases}$$
Multiplying both sides we get:
$$3^{m+r}-3^m\cdot 2^s-3^r\cdot 2^n +2^{n+s}=1$$
$$\big(\frac{3^m}{3^r}-\frac{2^n}{2^s}\big)2^s\cdot3^r+1=2^{n+s}+3^{m+r}$$
$$\Rightarrow \big(3^{m-r}-2^{n-s}\big)2^s\cdot 3^r +1=2^{n+s}+3^{m+r}$$
If $m-r=1$ and $n-s=1$ then we have:
$$2^s\cdot3^r +1=2^{n+s}+3^{m+r}$$
Substituting $s=a, r=b, n+s=c$ and $m+r=d$ we get the equation. Each of equations in system have similar solutions in $\mathbb Z$ which may give solutions of final equation, i.e  we may proceed reversely and find the solutions under constrains $m-r=1$ and $ n-s=1$; $m, n, r, s $ $\in \mathbb Z$.
