Seven more than a power of two

Let $k^{2} = w$ . By the given information, we can assume that $w$ is a multiple of $3$ and also a perfect square. The possible value of $w = \{ 9, 81, 729, \text{etc} \}$ . First, let $w= 81$ , then $k=9$ . Now, we are going to do some trial and error.

If $y = 1$ , then there's no positive integer solution for $x$ .

If $y= 2$ , then $x= 5$ .

By the given information, there's only two triplets of positive integer that satisfy the equation above. Then, we have $(x,y,k) = \{ (1,1,3) , (5,2,9) \}$ are the only solutions. The rest, you know what to do.

Solutions generator:

for x in range(10):
    for y in range(10):
        for k in range(10):
            a = 2 ** x
            b = 7 ** y
            c = k ** 2

            if a+b == c:
                print (x, y, k)

$2^5+7^2=9^2 \implies 2+5+7=16$