Note that $(t^2-p)(t^2+p)=qr$. If $t^2-p=1$ then $p=(t-1)(t+1)$, forcing $t=2,p=3$ and $qr=7$, which does not have a solution. Hence $t^2-p=q$ or $t^2+p=q$. In either case, $t^2\ge q-p$. (a). Let $n\ge 3$. Then $(p+t)^n\ge (p+t)^3 \ge p^3+t(3p^2+3pt+t^2)\ge p^3+t(3p^2+3pt+q-p)$. As $3p^2+3pt+q-p>q$ clearly, we get $(p+t)^n>p^3+qt>p^2+qt$, a contradiction. Hence $n\le 2$. (b). Notice that $n=2$ as $p^2+qt>p+t$. With this, we get $2p+t=q$ that is $q-2p=t>0$. Plugging this in, we obtain $p^2+qr = (q-2p)^4$. Inspecting both sides modulo $q$, we get $q\mid 16p^4-p^2 = p^2(4p-1)(4p+1)$. As $q>2p$, we get $q\mid (4p-1)(4p+1)$. Moreover, $2q\ge 4p+2>4p+1$ and thus either $q=4p-1$ or $q=4p+1$. The former case yields $p^2+(4p-1)r = (2p-1)^4$, yielding $r=(p-1)(4p^2-3p+1)$. As $r$ is prime and $4p^2-3p+1\ne 1$, we must have $p=2$. With this, we obtain $(p,q,r)=(2,7,11)$, $n=2$ and $t=3$. The case, $q=4p+1$, yields $r=(p+1)(4p^2+3p+1)$, thus $r$ is composite.