2024/052) Solve in integers $x^3+3367=2^n$

We are given $x^3+3367=2^n$

We know $3367=7 * 13 &*37

So let is work $x^3=2^n \pmod 7$

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

So we get

$x^3= 2^{3k} \pmod 7$

Going back to the original equation we get

$x^3 + 3367 = 2^{3k}$

Or $(2^k)^3 - x^3 = 3367$

Or $y^3 -x^3 = 3367$ where $y = 2^k$

As $(y-x) | y^3-x^3$ so $y-x |  3367\cdots(1)$

Further 

As we know $(y-x)^3 = y^3-3y^2x + 3yx^2 - x^3 = (y^3-x^3) - 3yx(y-x) \lt y^3-x^3$ when $y-x \gt 0$

So $y-x\lt 15\cdots(2)$

So $y-x \in \{1,7\}\cdots(2)$ from (1) and (2)  

 $y-x =1$

gives $x^3 + 3367 = (x+1)^3 = x^3 + 3x^2 + 3x + 1$

Or $3367 = 3x^2+ 3x + 1$

Or $3x^2+3x = 3366$

Or $x^2 + x = 1122$

x = 33 and y = 34 and y is not power of 2 so not a solution

$y-x=7$

gives

$x^3 + 3367 = x^3 + 21x^2 + 147 x + 343$

or $3024 = 21x^2+ 147x

or $x^2+ 7x = 144$

or $x = 9$ or $x=-16$

only positive value admissible giving x = 9 and y = 16

$2^n = y^3$ giving $2^n = 2^{12}$ or n = 12

Hence x=3 , n = 12