The next two numbers in the series may be 135 and 71 if the following reasoning is adopted.
There are 4 numbers in the series as given. They could be consecutive values of a polynomial of degree 3:
y=Ax^3+Bx^2+Cx+D, because there are 4 unknown coefficients and 4 numbers in the series, which is sufficient to find the unknowns.
Put x=0 and y=31, and we have D=31.
Put x=1 and y=47, and we have A+B+C+D=47, so A+B+C=47-31=16.
Put x=2 and y=91, and we have 8A+4B+2C+D=91 and 8A+4B+2C=91-31=60.
We can write: 2A+2B+2C=32 and subtract from the previous equation: 6A+2B=28, from which B=14-3A.
We know from above that C=16-A-B, so we can substitute for B and write C=16-A-14+3A=2+2A.
We now have B and C in terms of A.
Put x=3 and y=131, and we have 27A+9B+3C+D=131 and we can substitute for B, C and D:
27A+9*14-27A+6A+6=131. From this we get 6A=-32 and A=-16/3.
We can now find B and C from A: B=14+16=30; C=-26/3.
So y=-16x^3/3+30x^2-26x/3+31=30x^2+31-(2x/3)(8x^2+13).
Put x=4 and we get y=480+31-(8/3)(128+13)=511-8*47=135, the next term in the series.