2^a + 2^10 + 2^13 = b^2

Looking first just at 2^10 + 2^13, since 10 is less than 13 we can take out a factor of 2^10 to get:

2^10(1 + 2^3)

2^10 is obviously (2^5)^2 and (1 + 2^3) is 9, also a square number, therefore, you can replace 2^10 + 2^13 by (3*2^5)^2, or 96^2

2^a + 96^2 = b^2

Moving this to the right-hand side we have:

2^a = b^2 - 96^2

Factoring the difference of two squares gives:

2^a = (b+96)(b-96)

This means that both of the factors on the right-hand side are powers of 2, let’s call them 2^c = b+96 and 2^d = b-96.

If we take their difference to eliminate b we get:

2^c - 2^d = 192

Since d is smaller than c, we can take out a factor of 2^d:

2^d(2^(c-d)-1)=192

which is a product of a power of 2 and an odd number, which must therefore be 2^6 and 3.

d is therefore 6, and c is 8. a, as the sum of c and d, is therefore 14, and b is 2^6 + 96 = 160.

So the completed equation is

2^14 + 2^10 + 2^13 = 160^2