easy yet interesting.

hint : 2336 = 2²⁰+2⁸+2⁵

By writing p^k\mid \mid A, we mean exponent of p in A is k.

In the given equation x=y=z does not yield any solution, since 3\nmid 2336.

w.l.o.g \text{min}\left \{ x,y,z \right \}= z

So 2^z\mid \mid (2^x+ 2^y + 2^z) and 2^5\mid \mid 2336

so z=5 and we are left with 2^x + 2^y = 2304, Since 2304 is not a power of two , x=y does not give any solution, so we may assume x>y. Then 2^y\mid \mid (2^x + 2^y) and 2^8 \mid \mid 2304 so y=8, and we are left with 2^x = 2048 \implies x=11, so x=11,y=8,z=5 is the only solution...

Edit: sorry...there is one more case when min{x,y,z} = y= z which will not create too much of trouble... and

2^x + 2^y + 2^z = 2336
2336 = (2^5)(73).
so we have
2^(x-5) + 2(y-5) + 2^(z-5) = 73
rhs is odd. so we must have either x = 5 , or, y = 5 or z = 5.
take any of them equal to 5.
i take x = 5.
we are left with
2(y-5) + 2^(z-5) = 72
72 = (2^3)(9).
again use the odd - even thing...
and at last we get
(x,y,z) = all permutations of (11,8,5).