MHB What are the solutions to x^3+3367=2^n?

[anemone]

Here is this week's POTW:

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Determine all positive integers $x,\,n$ that satisfying the equation $x^3+3367=2^n$.

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[anemone]

Congratulations to kaliprasad for his correct solution to last week's POTW, which you can find below:

Spoiler

We are given $x^3 + 3367 = 2^n$
Now 3367 = 7 * 13 *37
so let us work mod 7
$x^3 = 2^n \pmod 7$

Working mod 7 we have $x^3 \in \{ 1,-1,0\}$ and $2^n \in \{2,4, 1\}$ so we get common as 2 and for that case n has to be multiple of 3 so say 3m

So we have $x^3 = 2^{3m} \pmod 7$

Going back to the original equation we have
$x^3 + 3367= 2^{3m} \pmod 7$
or $(2^m)^3 - x^3 = 3367$'

We can put $y= 2^m$ to keep it simple
So $y^3-x^3 = 3367$
as $(y-x) | (y^3-x^3$
so y = x = 1 or 7 or 13 or 37
as $(y-x)^2 = (y^2 + x^2 - 2xy) < y^2 + xy + x^2 $ so we have
$(y-x) = 1$ or 7 or 13

By checking y - x = 1 and 13 we do not get any root (method as as below but not mentioned and y-x = 7 gives root as y = 16, x = 9 as below

$ y - x = 7\cdots(1)$
so $y^2 + x^2 + xy = 481$
or $(y-x)^2 + 3xy = 481$
or $7^2 + 3xy = 481$
or $xy = 144\cdots(2)$
From (1) and (2) we get $y=16, x = 9$ or $n=12,x = 9$