How can I find out if two parabolas will share an integer in their sequence?

You are trying to solve $4y_1^2+48y_1+24 = 4y_2^2+4y_2$ for integer solutions. This can be rewritten as: $4(y_1+6)^2-(2y_2+1)^2=119$ or $(2y_1+2y_2+13)(2y_1-2y_2+11)=119$

All possible solutions now must be from one of the allowable factorisations: $119 = 1\times 119 = 7\times 17$, where ordering and $\pm1$ could be used to generate more cases. From these eight cases, we get all the solutions $(y_1, y_2) \in \{ (24, -30), (-36, 29), (24, 29), (-36, -30), (0, -3), (-12, 2), (0, 2), (-12, -3)\}$