Find all integer solutions of $3^a +7 = 2^b$ [duplicate]

I want to find all integer solutions of $$3^a + 7 = 2^b$$

I have found (by brute force) the two solutions
$$3^0 + 7 = 2^3$$ and
$$3^2 + 7 = 2^4$$

but I want to see if there are more solutions. I have found that if (b mod 3) = b' then (a mod 6) must be 2b', and that (a mod 4) can't be 1, but that's as far as I get and I have no clue how to make progress.
Any ideas? TIA

• I am pretty sure someone will find a complete proof, but $3^a+7$ is upto $a=10^5$ only a perfect power if $a=0$ or $a=2$ giving your solutions (brute force with PARI/GP) . It is very unlikely that there are more. Note that in general such diophantine equations are extremely difficult to solve or even out of reach , this particular one is probably not of this kind. Commented Apr 8, 2022 at 7:54
• – R.P.
Commented Apr 8, 2022 at 9:37
• Does this answer your question? Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$. Found from RP_'s comment above. Commented Apr 8, 2022 at 9:48

A negative $$b$$ or a negative $$a$$ cannot lead to a solution. So $$a,b\ge 0$$. For $$a=0$$ we obtain OP's first solution $$3^0+7=2^3$$. Else $$a\ge 1$$. Then taken modulo $$3$$ the powers of two ($$2\equiv -1\ [3]$$) for $$b=0,1,2,3,\dots$$ are $$1,-1,1,-1,\dots$$, so the involved two-power $$b$$ is even, $$b=2B$$ for some natural $$B$$.
Since $$b\ge 3$$, considering the L.H.S. modulo four, we have for $$a=0,1,2,3,\dots$$ the values $$1+7, -1+7, 1+7, -1+7,\dots$$ so the involved three-power is even, $$a=2A$$ for some natural $$A$$. We rewrite now the given relation as follows: $$7 = 2^a-3^b=2^{2A}-3^{2B}=(2^A-3^B)(2^A+3^B)\ .$$ This is possible only when $$(2^A-3^B)=1$$ and $$(2^A+3^B)=7$$, so $$2^A=(7+1)/2=4$$, and $$3^B=(7-1)/2=3$$, leading to the second solution given by the OP.
$$\square$$