Take the equation mod 3; then we get 1 = 2^z mod 3.
For this to be true then z is even; now let z=2w.
Then we have 64 + 3^x*5^y = 17^(2w)
3^x*5^y = 17^(2w) - 8^2
3^x*5^y = (17^w - 8) * (17^w+8)
In the last line the difference of factors is 16. For this to happen from a factorization of 3^x*5^y then all the factors of 3 are all in the same factor and also for all the factors of 5.
So the factorization is either {1, 3^x*5^y} or {3^x, 5^y}.
The first case implies 1+16=17 can be decomposed in the form 3^x*5^y, which is false so no solutions here.
Then we must have a power of 3 and a power of 5 which differ by 16. The only such pair I know of is 5^2-3^2 = 16.
x=2 and y=2 then means 64 + 3^2*5^2=17^z, which simplifies to 289 = 17^2 = 17^z.
Thus the triplet is (x,y,z)=(2,2,2).
Solution
