Determine all sets of non-negative integers x,y and z which satisfy the equation $2^x + 3^y = z^2$ 
Determine all sets of non-negative integers x,y and z which satisfy the equation $2^x + 3^y = z^2$

This came in the 1992 INMO and curiously enough also seems to have been included on the 1996 BMO Round 2? I have never heard of a question being directly copied from another Olympiad so this was a first for me.
Anyways, first, I looked at the case $y=0$. This quickly gave me one solution, viz $(x,y,z)=(3,0,3)$
Next, I considered $x,y,z>0$
We know $2^x + 3^y \equiv (-1)^x+0 \bmod 3$ and that perfect squares are $\equiv 0,1 \bmod 3$. It is easy to see that the only combination that works is $x$ be even and $z=3m+1$ type $\Rightarrow z$ is odd
Also, we know that odd perfect squares are $\equiv 1 \bmod 4$. Further, $3^y\equiv (-1)^y \bmod 4$ and since $x$ is even it implies that $x≥2$ thus $2^x$ is divisible by $4$. This further implies that $(-1)^y \equiv 1 \bmod 4 \Rightarrow y$ is also even.
Let $x=2k$. Then our original expression becomes $$3^y=(z+2^k)(z-2^k)$$
We have two possibilities: first is that $(z-2^k)=1$ and $(z+2^k)=3^y$ and second is $(z-2^k)=3^{y-a}$ and $(z+2^k)=3^a$. But since we previously established that $z=3k±1$ and as $2^k \equiv (-1)^y \bmod 3$, we can quickly discard the second possibility.
So we finally have, $$(z-2^k)=1$$ $$(z+2^k)=3^y$$
Here I got woefully stuck. Another thing I got was that $k$ is also even (which means $x$ is itself a multiple of $4$). One more thing is that since $y$ is even $3^y$ is divisible by $9$. I don't know how we can use this fact right now but I thought it could be worth mentioning.
Any help to proceed would be appreciated, thanks.
 A: First, there are a few minor problems with your proof:

Next, I considered $x,y,z>0$

Have you found all solutions with $xyz=0$? (No!)

We know $2^x + 3^y \equiv (-1)^x+0 \bmod 3$ and that perfect squares are $\equiv 0,1 \bmod 3$. It is easy to see that the only combination that works is $x$ be even and $z=3m+1$ type $\Rightarrow z$ is odd.

It is true that $x$ must be even, but not that $z\equiv1\pmod{3}$. It is also possible that $z\equiv2\pmod{3}$. Fortunately you later state that $z=3k\pm1$, so perhaps this is just a typo. But the conclusion that $z$ is even seems out of place; instead this follows from the simple fact that $x>0$, as then
$$z^2\equiv 2^x+3^y\equiv1\pmod{2}.$$
The rest of the proof is fine. The linked duplicates provide alternative (and quicker) solutions to your original problem, but here's a quick and easy finish to your approach:
You already note that $y$ is even, so
$$2^{k+1}=(z+2^k)-(z-2^k)=3^y-1=(3^{y/2}+1)(3^{y/2}-1).$$
Then both factors on the right hand side are powers of $2$, and they differ by $2$, so $y=2$.

As noted in the comments, this is a special case of Mihăilescu's theorem, previously known as Catalan's conjecture. It was still a conjecture at the time these questions were posed in the IMO contests, so it's safe to say you were not expected to know or use Mihăilescu's theorem. Participants with an interest in number theory might be aware of the conjecture (it's quite famous), so at least they would 'know'  that this should be the only solution.
