Two other solutions so far and they both used a brute-force computer for part 2. You don't need that massive overkill.

The part 2 equation is 2xy+3x+2y=N.

Add 3 to each side and factor (x+1) * (2y+3) = N+3

x and y are to be nonnegative, then x>=0 implies x+1>=1 and y>=0 implies 2y+3>=3.

The first conclusion is that the factorization of N+3 must be into two positive integers. But for 2y+3=(factor) to have a nonnegative integer solution that factor must be an odd number larger than 1.

Thus N+3 must have a nontrivial odd factor. This describes all integers that are not a power of 2. So the answer for part 2 is all integers except the integers that are of the form N=2^x-3; specifically **all integers in [10,99] except N=13, 29, and 61** may be used in the equation and there will still be nonnegative solution.

With that, I'll backtrack to part 1. N=42 makes N+3=45. Let N+3=f*g with x+1=f and 2y+3=g. The factorizations of 45 with g being odd and larger than 1 are (1,45), (3,15), (5,9), (9,5), and (15,3). Then **(x,y) = (0,21), (2,6), (4,3), (8,1), or (14,0)**.