<s>[INST] 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?
 [/INST] 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 :
$$
(2,0,1,1),\; (4,0,3,2), \;(-2,2,-2,1),\; (2,-1,1,-1).
$$
</s>