There is an exercise in Ch X of TY Lam's book Introduction to quadratic forms which asks us to prove that if a 5-dimensional quadratic form is universal then it is isotropic.
I have been able to prove an easier version of it which shows if a 3-dimensional quadratic form is universal then it is isotropic.
I approached it in the following way: Let $\sigma$ be a 5-dim form with d($\sigma$)=d. Since $\sigma$ is universal , it represents d . Hence $\sigma$ $\cong$ $<d>$ $\bot$ $\gamma$ where $\gamma$ is 4- dim form with d($\gamma$)=1. Hence, $\gamma$ is a Pfister neighbor and a.$\gamma$ $\subseteq$ $\tau$ where $\tau$ is 2-fold Pfister form.
I am stuck here and don't know how to proceed from here.
Your attempted solution is the correct way to start. Let's now finish your solution using the notation you have already introduced. We have $a \sigma = <ad> \perp a \gamma = <ad> \perp \tau$, where $\tau$ is a $2$-fold Pfister form. Let $\psi = <1, ad> \otimes \tau$, which is a $3$-fold Pfister form. Since $a\sigma$ is also universal and is a proper subform of $\psi$, it follows that $\psi$ is isotropic. Since $\psi$ is a Pfister form, it follows that $\psi$ is hyperbolic. Any $5$-dimensional subform of $\psi$, such as $a\sigma$, must be isotropic. Therefore, $\sigma$ is isotropic.