$x^2+xy+y^2=7^n~~~~~~~~~~~~~~~~~~(1)$
In its present wording the problem does not require to find out the triplets $~(x,~y,n)~$ ,

but rather the powers of seven which can be represented as $~~x^2+xy+y^2,~~$ with the proviso that

$~7\nmid xy~~$

It can be derived from the literature regarding $~~x^2+xy+y^2=N~~$ that if $~(x,y)=(a,b),~(c,d)~~$ are

two solutions to $~x^2+xy+y^2=7^n,~~~(x,y)=(ac+bc+bd,~ad-bc)~~$ is a further solution by virtue

of the multiplicative property of numbers under the form $~~x^2+xy+y^2=N~~$ (see (*) below)

Therefore, it is sufficient to depart from two solutions, one selected from the group consisting of:

$x^2+xy+y^2=7~~\Longrightarrow (a,b)=(1,2), (2,1), (-1,3), (-2,3), (-3,2), (-3,1), (-2, -1), (-1,-2), (1,-3), (2, -3), (3,-2),$

$(3, -1)$

and the second one selected from the group consisting of:

$x^2+xy+y^2=49~~\Longrightarrow (c,d)= (3,5), (5,3), (-3,8), (-5,8), (-8,5), (-8,3), (-5,-3),(-3,-5), (-8,3), (8,-5)$

to derive further solutions corresponding to any power of seven by applying the formula:

$(x,y)=(ac+bc+bd,~ad-bc)$

Of course, care should be taken that $~~7\nmid ac+bc+bd,~ad-bc~~$

Example 1

$a=1, b=2, c=3, d=5\Longrightarrow (1*3+2*3+2*5=19,~~1*5-2*3=-1)\Longrightarrow 19^2-19+1=343=7^3$

Example 2

$a=-3, b=1, c=19, d=-1\Longrightarrow (-39,-16)\Longrightarrow 39^2+39*16+16^2=2401=7^4$

Example 3

$a=8, b=-5, c=19, d=-1\Longrightarrow (62,~87)\Longrightarrow 62^2+62*87+87^2=16807=7^5$

And so on.

It might be possible to generalize this approach to the power of other primes (at first sight 3 and 13 could

work).

(*)Click to reveal hidden text