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?

There might not be an easy elementary solution to this equation. It was solved (there are $12$ nontrivial solutions in integers, where trivial solutions have at least $2$ of the exponents equal to $0$) by Tijdeman and Wang in a paper in the Pacific Journal in 1988; their argument uses bounds for linear forms in logarithms. The nontrivial solutions are as noted by Robert Israel, as well as $(a,b,c,d)$ in the following list :
$$
(2,0,1,1),\; (4,0,3,2), \;(-2,2,-2,1),\; (2,-1,1,-1).
$$
