Track coefficients in a composite binomial

Put $u=px-3x^2$. Then $(1+u)^6=1+6u+15u^2+20u^3+15u^4+6u^5+u^6$. The coefficient of $x^2$ is $6(-3)+15p^2=-18+15p^2=42$, so $p^2=4$ and $p=\pm2$.

The relevant terms are $15(px-3x^2)^2=15[p^2x^2-6px^3+9x^4]$, $20(px-3x^2)^3=20[p^3x^3+3p^2x^2(-3x^2)+\cdots]$ and $15(px-3x^2)^4=15[p^4x^4+\cdots]$. Thus the coefficient of $x^4$ is $135-180p^2+15p^4=135-720+240=-345$. Powers $u^5$ and $u^6$ cannot contribute to $x^4$.