For which integers x, y is $2^x + 3^y$ a square of a rational number? For which integers x, y is $2^x + 3^y$ a square of a rational number?
(Of course $(x,y)=(0,1),(3,0)$ work)
 A: Case 1:
Let's try the case $x,y\geq 1$ first. Let $2^x+3^y=n^2$ (where $n>0$). We notice that $3\not|n$, therefore $n^2\equiv 1(\mod\,3)$. Now consider the equation $2^x+3^y=n^2$ $(mod\,3)$, we get that $(-1)^x\equiv n^2\equiv 1(\mod\,3)$. Hence $2|x$. Let $x=2u$. Thus, $4^u+3^y=n^2$. Now move the $4^u$ to the other side to get a difference of squares:
$$3^y=n^2-4^u=(n-2^u)(n+2^u)$$
3 can't divide both of $(n-2^u),(n+2^u)$. Hence either $n+2^u=1$ (clearly not the case) or $n-2^u=1$.Thus, $n+2^u=3^y$. Hence:
$$1=n-2^u=3^y-2^{u+1}$$
Now we know from Catalan's Conjecture that ..... (http://en.wikipedia.org/wiki/Catalan's_conjecture) and we're done with the first case.
I am almost sure the other cases can be handled similarly but I feel lazy (I might be wrong though).
Edit: Ivan Loh noted that Catalan's Conjecture may be avoided. The argument that would replace Catalan's conjecture is in the comments
A: There are also negative integer solutions as well.  Consider $(-4,-2)$ where the expression then becomes $\frac{1}{16}+\frac{1}{9} = \frac{9}{144}+\frac{16}{144} = \frac{25}{144}$ which is the square of $\frac{5}{12}$.
