We have given three positive integers $p,q,r$ satisfying the following conditions: $(1) \;\; r = q^4 - p^4$, $(2) \;\; p,q \in \mathbb{P}$, $(3) \;\; \tau(r) = 8$. Clearly $p<q$ by condition (1), which means $q$ is odd by condition (2). Assume $p \neq 2$. Then $p$ is odd, which implies $16 \mid r = q^4 - p^4 = (q^2 - p^2)(q^2 + p^2)$ since $8 \mid q^2 - p^2$ and $2 \mid q^2 + p^2$. If $r$ has an odd prime divisor $t$, then $2^m \cdot t^n$, where $0 \leq m \leq 4$ and $0 \leq n \leq 1$, are divisors of $r$, i.e. $r$ has at least 10 positive divisors, which contradicts condition (3). Consequently $r$ is a power of 2, yielding $r = 2^7$ by condition (3). Therefore by equation (1) $(4) \;\; (q^2 - p^2)(q^2 + p^2) = 2^7$. Now $4 \mid q^2 + p^2 - 2$ implies $q^2 + p^2 \leq 2$ by equation (4), which obviously is impossible. Hence $p=2$, which inserted in equation (1) result in $(5) \;\; r = (q - 2)(q + 2)(q^2 + 4)$. If $q=3$, then $r = 3^4 - 2^4 = 81 - 16 = 65 = 5 \cdot 13$, yielding $\tau(r) = 4$. Hence $q \neq 3$ by condition (3). The positive integers $q - 2$, $q + 2$ and $q^2 + 4$ are coprime, which according to condition (3) and eqution (5) implies $(6) \;\; \tau(q - 2) \cdot \tau(q + 2) \cdot \tau(q^2 + 4) = 8$. Now $q^2 + 4 > q + 2 > q - 2 > 1$ (since $q>3$), implying $\rho(q - 2), \rho(q + 2), \rho(q^2 + 4) \geq 2$, which combined with equation (6) give us $(7) \;\; \tau(q - 2) = \tau(q + 2) = \tau(q^2 + 4) = 2$. Hence $q - 2, q + 2$ and $q^2 + 4$ are three primes. Hence, since $3 \mid q \pm 2$, we obtain $q \pm 2 = 3$, yielding $q = 3 + 2 = 5$, which inserted in equation (5) result in $r = 3 \cdot 7 \cdot 29 = 609$ and $\tau(r) = 2^3 = 8$. Conclution:. There are only one triple $(p,q,r)$ of positive integers satisfying conditions (1)-(3), namely $(p,q,r) = (2,5,609)$.

Since $r$ has $8$ positive divisors, $r=a^7$ or $r=a^3b$ or $r=abc$, where $a,b,c$ are primes. Factorising, $r=q^4-p^4=(q-p)(q+p)(q^2+p^2)$. Obviously $q>p,\gcd(q,p)=1$. Also, what we'll refer to as the "factor gcd's": $\gcd(q-p,q+p)\mid 2p\implies \gcd(q-p,q+p)\in\{1,2\}$. $\gcd(q-p,q^2+p^2)=\gcd(q-p,(q-p)(q+p)+2p^2)\mid 2p^2\implies\gcd(q-p,q^2+p^2)\in\{1,2\}$. $\gcd(q+p,q^2+p^2)=\gcd(q+p,(q-p)(q+p)+2p^2)\mid 2p^2\implies\gcd(q+p,q^2+p^2)\in\{1,2\}$. If $p=2$ then $(q-2),(q+2),(q^2+4)$ are all odd so the factor gcd's are each $1$. So we must be in the case $r=abc$. But if $q-2$ and $q+2$ are prime, then $q=5$ or $q>5,q\equiv 0 (3)$ which is impossible. So $p,q,r=\boxed{2,5,609}$ in this case. Otherwise $p\ge 3$ and $(q-p),(q+p),(q^2+p^2)$ are all even so the factor gcd's are each $2$ and we must be in the case $r=2^3b$, which is clearly absurd since $(q-p),(q+p),(q^2+p^2)$ are all distinct. So there are no more solutions.