The question boils down to:
How many solutions are there for #x^2+y^2 = 4^2# ?
There are #4# solutions with #x, y in ZZ# namely #(0, 4)#, #(0, -4)#, #(4, 0)#, #(-4, 0)#
There are an infinite number of solutions with #x, y in RR#, namely
#(4cos theta, 4sin theta)# for #0 <= theta < 2pi#
How about solutions with #x, y in QQ# ?
For any non-negative integer #k#, define
#a_k = 2k+3#
#b_k = (a_k^2-1)/2 = 2k^2+6k+4#
#c_k = (a_k^2+1)/2 = 2k^2+6k+5#
Then #a_k#, #b_k# and #c_k# are the lengths of the sides of a right angled triangle and they are integers, e.g. #(3,4,5)#, #(5,12,13)#, #(7,24,25)#.
Then the point #((4a_k)/c_k, (4b_k)/c_k)# satisfies #x^2+y^2=4# and #x, y in QQ#.
Since there are a countably infinite number of choices for #k#, there are an infinite number of points #(x, y)# with rational coordinates at a distance of #4# from the origin.
