An efficient way to solve this.

Take the logarithm and treat as linear equation.

A^x+B^y+C=Z

becomes

xLog(A)+yLog(B)=Log(Z-C)

Which is recognized as the standard form of a line.

Caveat--problems arise when C>=Z.

Rap

log(a+b) != log(a) + log(b)

oops. disreguard

only Z is given. Am trying to find the right combination to get Z.

Do you only need a single solution? Are all variables integers?

Are you looking for a computer algorithm or a way to force a solution by hand? Your problem is woefully under specified. If you set A=B=0 and C=Z, that will always work, but if that was ok you wouldn't be here. There has to be something you aren't telling us.

C is constrained to be potentially much smaller than Z. The problem is trivial for small enough Z. It's when Z gets really large that it gets difficult. The problem seems to be to find integer powers of two (up to) 32-bit integers such that their sum is within 2^64 of an arbitrarily large (up to 2^8000) number.

For the vast majority of allowable Z values, there will not be a solution. There are fewer than 2.43E43 possible values that can be generated by the left hand side (all valid combinations of A, B, C, x, and y), but there are 1.7E2408 (2^8000) possible values for the right hand side (Z). You've only got a 1 in 7E2364 chance of picking a Z with a solution...

Assuming Z > 2^64 (otherwise it's trivial) and A^x >= B^y, then

Z > A^x >= (Z - 2^64)/2

If Z is quite large, for each A, there is only one x that satisfies the above inequality (although for A=2, there are potentially two values for x). That narrows the search, but you still have to check 2^32 values for A, and for each A, you need to do a similar thing to find a B^y that gets you within 2^64 of Z.

That's the best I can do.