$ x > 0$ or $ y > 0$ so $ 2^x \equiv 1 \bmod{3}$ or $ 2^x \equiv 1 \bmod{5}$, so $ x$ is even.
We can also check that $ n \ge 3$ so $ 3^y5^z \equiv 1 \bmod{8}$, therefore $ y$ and $ z$ are both even.
Then $ 2009 = 3^y5^z - 2^x = a^2 - b^2 = (a - b)(a + b)$
The factorisations of 2009 are $ 2009 = 1.2009 = 7.287 = 41.49$
Checking each case, the only factorisation that gives a solution is $ a + b = 49, a - b = 41 \implies a = 45 = 3^2.5^1, b = 4 = 2^2$
So the only solution is $ (x,y,z) = (4,4,2)$