$$(x,y) \in \lbrace(4,14),(3,2),(4,-14),(3,-2)\rbrace$$
Indeed, it's a difference of two squares:
$$2^{2x}-y^2=60$$
$$\text{let} \ \ \ \ t=2^x \ \ \implies $$
$$t^2-y^2= 60$$
In using the identity:
$$\left(\frac{d_1+d_2}{2}\right)^2-\left(\frac{d_1-d_2}{2}\right)^2=d_1\cdot d_2$$
the only catch is that the pairs of divisors you choose need to be of the the same pairity, to allow that their sum or difference is divisible by two. Moving on, we have:
$$t=\frac{d_1+d_2}{2} \qquad \text{and} \qquad y= \frac{d_1-d_2}{2}$$
The total number of divisors of a given natural N will be:
$$\text{if} \ \ \ \ N=\prod_{i=1}^M p_i^{\alpha_i} \ \ \ \ \text{for finitely large M, and the i-th prime p}$$
$$\text{then} \ \ \ \ \left| \lbrace \text{set of divisors of N} \rbrace\right|=\prod_{i=1}^M (\alpha_i +1)$$
So for any number you factor it completely, and multiply the powers increased by one to get the amount of divisors of that number.
We now observe that
$$60=2^2 \cdot 3 \cdot 5$$
$$(2+1)(1+1)(1+1)=12$$
So the divisors of 60 must include 12 numbers, starting from 1 it's not hard to find that they are:
$$\lbrace 1,2,3,4,5,6,10,12,15,20,30,60 \rbrace$$
And that the set of divisor-pairs, with each divisor of the same parity is only a two member set, so:
$$(d_1,d_2)\in \lbrace(30,2),(10,6)\rbrace$$
Thus
$$t=2^x=\frac{30+2}{2} \ \ \ \lor \ \ \ \frac{10+6}{2}$$
$$\text{so} \ \ \ \ t=16 \ \ \ \ \lor \ \ \ 8 \ \ \ \implies x=4 \ \ \ \lor \ \ \ 3$$
While
$$y=\frac{30-2}{2}\ \ \ \lor \ \ \ \frac{10-6}{2} \implies y=14 \ \ \lor \ \ \ 2$$
Answers are then
$$(x,y)=(4,14) \lor (3,2)$$
But we can see that we can let y be positive or negative, and not so for x, giving our above answer of 4 ordered pairs