$ 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)$

x should be A MULTIPLE OF 4 BECAUSE RIGHT HAND SIDE MUST BE A MULTIPLE OF 5 NOW ON LEFT HAND SIDE WE HAVE A ODD NUMBER + AN EVEN NUMBER WHICH CAN BE A MULTIPLE OF 5 ONLI WHEN THE FINAL NUMBER ON LEFT HAND SIDE HAS A LAST DIGIT OF 0 OR 5 YA IT SHOULD BE ODD BUT IT SHOULD ALSO BE A MULTIPLE OF 5 SO THE ONLI WAY TO MAKE THE FINAL LHS NUMBER TO END IN 0 OR 5 IS FOR 2 raised to the power TO HAVE THE LAST DIGIT AS 6 SO WE HAVE 1 2 4 16 32 34 128 256 AND SO ON 16=2^4 256=2^8 AND SO ON NOW U TRY X=4 THAT GIVES U 2009+16 =2025 =3^4*5^2 SO CASE 1: X=4,Y=4,Z=2 so the onli way is that the number on LHS must end with a 5