this is my solution so far

We are looking for solutions of:

2^x + 3^y = z^2 [1]

where x, y, and z are non-negative integers.

So z^2 ≡1 (mod 3) unless z^2=3m for any integer m, and if x is odd, say x=(2k+1) for some integer k. However when z=3, we have 2^3+3^0=3^2, so one solution is when x=3,y=0 and z=3 When x is odd 2^(2k+1)≡2(mod 3) and when x is even 2^x≡1 (mod3) so z^2 is congrgent to 2^x when x is even.

We now know x is even, let x=2k. So we have:

3^y=(z-2^k )(z+2^k )

Now (z-2^k ) and (z+2^k ) should both be powers of three, so therefore (z+2^k )-(z-2^k ) should equal a power of three, so

3^w=(z+2^k )-(z-2^k )

3^w≠2^2k=2^x

So 2^x should be divisible by three, but it isn’t. So let (z-2^k )=3^a and (z+2^k )=3^b then we have

█((z+2^k )=3^b@(z-2^k )=3^a )/(2^(k+1)=3^b-3^a )

If a>0 then 3^b-3^a will be a power of three and 2^(k+1) is not divisible by three, so let a=0. We are left with

2^(k+1)=3^b-1

Let a=k+1, then we have 2^a=3^b-1.So this means that b≥0, otherwise 2^a wont be an integer, which means a≥1. Also a≤3 because if k>2, then 3^b+1≡0mod(8) which is a contradiction, so therefore a has to be either 1,2 or 3, making k=0,1 or 2.

When k=0 we have

2^1=3^b-1 → 3=3^1 so therefore b=1 and

When k=1 we have

2^2≠3^b-1 so therefore k=1 doesn’t exist.

When k=2 we have

2^3=3^b-1 → 9=3^b so therefore b=2.

So the solutions of 2^x+3^y=z^2 are

X 0 4 3

Y 1 2 0

Z 2 5 3