Ordered pairs of positive integers $2^{2x} - y^2 = 60$ How many ordered pairs of positive integers are there that satisfy $2^{2x} - y^2 = 60$? 
This can be rewritten as $(2^x)^2-y^2 = 60$ and then $(2^x+y)(2^x-y) = 60$. Then since $2^x$ is always positive and so $2^x+y$ and $2^x-y$ are both positive and the first is always bigger than the latter. This means I only have to account for the positive factors of $60$, and so I had the following pairs $(60,1), (30,2), (20,3), (15,4), (12,5)$, and I plugged them in and solved the systems. Of the 5 possible cases, only $(30,2)$ worked and yielded $x=4, y=14$. 
Am I right in having done so? They don't have an answer, and when I tried to graph this, it didn't exactly work out well (I didn't get an integer value for $x$). 

No you are not right.
You have neglected the case $\{ 2^x-y, 2^x+y \} = \{ 6, 10 \}$ 
; which gives you $(x,y)=(3,2)$.


Notice that: 


*

*$(2^x+y)$ and $(2^x-y)$ have the same pairity; 

*product of $(2^x+y)$ and $(2^x-y)$ is even; so at least one of them is even;  

*by the above two remarks it follows that 
both of $(2^x+y)$ and $(2^x-y)$ are even.  



*

*So we can rewrite the equation as 
$\dfrac{2^x+y}{2} \cdot \dfrac{2^x-y}{2} = 15$ ; 

*the only possibilitis for $\{ \dfrac{2^x-y}{2}, \dfrac{2^x+y}{2} \}$ 
is $\{ 1, 15 \} $ and $\{ 3, 15 \} $ . 

*So we can conlude that:
the only possibilitis for $\{ \ 2^x-y, \ 2^x+y \ \}   $ 
is $\{ 2, 30 \} $ and $\{ 6, 10 \} $ .
The first possibility gives $(x,y)=(4,14)$ 
and the second gives you $(x,y)=(3,2)$ . 
