I can verify solutions to my problem in polynomial time, how would a non-deterministic algorithm arrive to a solution if it always takes $2^n$ bits?

Decision Problem: Given integers as inputs for $ K$ and $ M$ . Is the sum of $ 2^k$ + $ M$ a $ prime$ ?


m = int(input('Enter integer for M: ')) sum_of_2**K+M=int(input('enter the sum of 2^k+m: '))  if AKS.sum_of_2**K+M == True:    # Powers of 2 can be verified in O(N) time   # make sure there is all 0-bits after the 1st ONE-bit      # Use difference to verify problem    if sum_of_2**K+M - (M) is a power_of_2:     OUTPUT Solution Verified 

The powers of 2 have approximately $ 2^n$ digits. Consider $ 2^k$ where $ K$ = 100000. Compare the amount of digits in $ K$ to the amount of digits in it’s solution! Also take note that the powers of 2 have $ 2^n$ bits as its 0-bit Unary essentially for the exponent $ n$ .


How would a non-deterministic machine solve this problem in polynomial time?