Find all positive integers $ a $ and $ b $ such that $7^a-3(2^b)=1$ I am looking for all positive integers satisfying the equation
$$7^a-3(2^b)=1$$
I used the fact that $ (1,1) $ is a solution and got
$$(7^a-7)=3(2^b-2)$$
I thaught about using Gauss Theorem in vain.
Thanks in advance for any idea.
 A: Hints:

If $b\geq 3$ then $$(-1)^a \equiv_8 1 \implies a \equiv_2 0$$ so $a=2c$ and now we have $$(7^c-1)(7^c+1)=3\cdot 2^b$$


 Clearly, since $7^c-1$ and $7^c+1$ are two consecutive even numbers exactly one is not divisible by $4.$

Hope you can finish now.
A: An argument mod $7$ shows that $b+2$ is divisible by $3$, then let $b=3n-2.$
We can take the two cases $a=2m, a=2m+1.$
The problem can be reduced to finding the integer points on elliptic curves as follows.
$\bullet\ a=2m$
Let $X=3\cdot2^{n}, Y=6\cdot7^{m}$, then we get
$Y^2 =X^3 + 36.$
According to LMFDB, this elliptic curve has integral solutions $(X,Y)=(-3,\pm 3), (0,\pm 6), (4,\pm 10), (12,\pm 42).$
From $(12,\pm 42)$ we get $(m,n)=(1,2) \implies (a,b)=(2,4).$
$\bullet\ a=2m+1$
Let $X=21\cdot2^{n}, Y=294\cdot7^{m}$, then we get
$Y^2 =X^3 + 12348.$
According to LMFDB, this elliptic curve has integral solutions $(X,Y)=(-14,\pm 98), (-3,\pm 111), (21,\pm 147), (37,\pm 251), (42,\pm 294), (378,\pm 7350), (11802,\pm 1282134).$
From $(42,\pm 294)$ we get $(m,n)=(0,1) \implies (a,b)=(1,1).$
Hence there are only integral solutions $(a,b)=(1,1),(2,4).$
