Unless I am mistaken, no solutions. Checking $\pmod{7}$, one can verify $3|n=3m$, say. Then $(2^m-x)(2^{2m}+2^mx+x^2)=3367=7\cdot 13\cdot 37$. Let $2^m-x=a$. Now $a<2^{2m}+2^mx+x^2=\frac{3367}{a}\implies a<59<7\cdot 13$ so $a\in\{1,7,13,37\}$. Now $a((x+a)^2+(x+a)x+x^2)=3367\implies 3ax^2+3ax+a^3-3367=0$, the discriminant being $12\cdot 3367a+9a^2-12a^4$, which must be a perfect square, but sadly isn't for $a\in\{1,7,13,37\}$.

x=9 and n=12 is a solution.I think you can delete the 37 using a stronger inequality and then just try the other cases

Woops - slight mistake. The quadratic should be $3ax^2+3a^2x+a^3-3367=0$. You're right about using a stronger inequality for $a$ to ignore the case $a=37$. Note $a^2=(2^m-x)^2<2^2m+2^mx+x^2=\frac{3367}{a}$ and hence $a^3<3367\implies a< \sqrt[3]{3367}\approx 14.988$. So $a\in\{1,7,13\}$. Frankly, from here I would just substitute the possible values of $a$ into the quadratic rather than deal with the large discriminant. For $a=1$ we obtain the quadratic $x^2+x-1122=0$, so $x=33$, however $2^m=x+a=34$, impossible. For $a=7$, we obtain the quadratic $x^2+7x-144=0$ so $x=9\implies n=12\implies(12,9)$. For $a=13$, we get $x^2+13x-30=0$ so $x=2$, no solution. So only solution is $(12,9)$. In fact here is another method: $3|n$ as mentioned before, looking at $\pmod{9}$ then shows $2|n$ thus $6|n$. Now $2^6<3367$ so $n\not= 6$. $n=12$ gives $x=9$. Otherwise $n\ge 18$. But the difference between $2^{18}$ and it's nearest cube, $(2^6-1)^3$ is $12097>3367$ so when $n$ is that large, cubes are too spaced out for any difference between them to be $3367$. Hence $n=12,x=9$ is the sole solution.

Looking at the equation $mod 7$ we get that n is a multiple of 3 . Let $n = 3k$ for some positive $k$ than we have $(2^k - x)(2^{2k } + 2^kx + x^2)$ = 3367. Checking the positive multiples of 3367 for this gives only $n=12, x = 9 $ as a solution.