(x,y) = (4, 6) constitutes the only possible solution to the equation under reference.
EXPLANATION:
Let, if possible, at least one of x and y be odd.
If x is odd, then 2^y = (1)^y(Mod 3)= 1(Mod3)
or, 2^y  3^x = 1(Mod 3).
However, 2^y  3^x =  17= 1(Mod 3), which is a contradiction.
Accordingly, x must be even.
If y is odd, then 3^x= 3 (Mod 8).
Consequently, for y >=3, we observe that:
2^y  3^x = 3(Mod 8)= 5(Mod8).
By the problem, 2^y  3^x =  17
or, 2^y  3^x = 7 (Mod 8). This is a contradiction.Hence, if y is odd, it follows that y must be less than 3; or, in other words, y is equal to 1. But, for y=1, we obtain 3^x = 19, which is not feasible.
Consequently, both x and y must be even, so that:
x=2m and y=2n, for two positive integers m and n.
Accordingly, we obtain:
3^(2m)  2^(2n) = 17
or, (3^m + 2^n)(3^m  2^n)= 17*1
or, (3^m + 2^n, 3^m  2^n) =(17,1); (17, 1); (1, 17);(1, 17)
or, (3^m,2^n)=(9, 8)(9, 8); (9, 8); (9, 8);
Or, (m,n)=(2,3) corresponding to (3^m,2^n)=(9,8) and each of the other pairs for (3^m, 2^n) contains at least one negative integer, which is a contradiction, so that:
(x,y) = (4,6) which constitutes the only possible solution to the equation under reference.

For a different methodology leading to the same result, refer to the solution submitted by Dennis in this location.
