Inspired By Sophie Germain

Relevant wiki: Diophantine Equations - Solve by Factoring

We can factor the right side of the equation. $x^6+27=(x^2+3)(x^2-3x+3)(x^2+3x+3).$ Since $y$ is the product of two prime numbers, then at least one of the three factors must be $1$ or $-1.$ The first factor and the third factors can never be $1$ or $-1$ for any positive integer $x$ . The second factor can be $1$ when $x=1$ or $x=2,$ but it cannot be $-1$ for any positive integer value of $x.$ Therefore, the only possible values of $x$ are $x=1$ or $x=2.$ When $x=1,$ the value of the $y$ is 28, which is not the product of two primes, and when $x=2,$ the value of $y$ is $91,$ that is the product of $7$ and $13.$ Hence, the only solution of this equation is $(2, 91)$ , and then the answer to the question is $2+91=\boxed{93}.$

By Fermat's Little Theorem, we have the following:

$x^7 \equiv x \mod{7}$

Or, equivalently:

$x^6 \equiv 1 \mod{7}$

$x^6 + 27 \equiv 28 \mod{7} \equiv 0 \mod{7}$

So $7$ must be one of the prime factors.

$x^6 + 27 \equiv (x^2+3)(x^2-3x+3)(x^2+3x+3)$

None of the factors yield real solutions for $x$ when we set them equal to $-7$ .

The solutions when set equal to $7$ yield $1, 2, 4$ , after omitting negative solutions.

Case testing shows $x=2$ is the only valid solution.

So we have $(x,y) = (2,93)$ and the solution follows.