Suppose integer $m$ has $\phi(m)=2pq^5r^2$ where $p,q,r$ are primes.
Hellman-Pohlig says that finding discrete log $z\bmod p$, $z\bmod q^5$, $z\bmod r^2$ and $z\bmod 2$ suffices to find $z\bmod\phi(m)$ in $g^z=h\bmod m$.
It could be that $p,q^5,r^2$ are still very large and there is a Big-step-small-step which is still exponential.
Can we apply sieve techniques to each of $z\bmod p$, $z\bmod q^5$, $z\bmod r^2$ and $z\bmod 2$ cases separately?
I'm not sure if I understand the question correctly, but let's try anyway.
By assumption we have some integer $m$ such that $\varphi(m)=2pq^5r^2$ for primes $p,q,r$. The goal is to solve a discrete logarithm problem in $\mathbb{Z}_m^*$, say we have $g,h\in\mathbb{Z}_m^*$ such that $h=g^\ell$ for some integer $\ell$.
We note that $\mathbb{Z}_m^*$ is a finite abelian group of order $\varphi(m)$, so by the fundamental theorem of finite abelian groups we have that $\mathbb{Z}_m^*\cong\mathbb{Z}_2\times\mathbb{Z}_p\times G(q^5)\times G(r^2)$, where $G(q^5)$ and $G(r^2)$ are groups of order $q^5$ and $r^2$ respectively (they are Sylow subgroups). Pohlig-Hellman implicitly makes use of this fact. That is, instead of finding $\ell\pmod{\varphi(m)}$ directly, we map $g$ and $h$ to elements of $\mathbb{Z}_2$, $\mathbb{Z}_p$, $G(q^5)$ and $G(r^2)$ respectively. We then solve the DL's in those groups, finding $\ell\pmod{2}$, $\ell\pmod{p}$, $\ell\pmod{q^5}$ and $\ell\pmod{r^2}$. Finally we recombine them using the Chinese Remainder Theorem.
For example, since $g$ and $h$ have order dividing $\varphi(m)$, we have $g^{2pq^5r^2}\equiv h^{2pq^5r^2}\equiv 1\pmod{m}$. That means that $(g^{pq^5r^2})^2\equiv (h^{pq^5r^2})^2\equiv 1\pmod{m}$, i.e. they are both elements of order 2. In other words, we can think of them as lying in $\mathbb{Z}_2$. Thus, we have that $h^{pq^5r^2}=(g^{pq^5r^2})^{\ell\pmod{2}}$. This is an easy DL and we solve it to find $\ell\pmod{2}$.
So we have reduced to solving DL's in $\mathbb{Z}_2$, $\mathbb{Z}_p$, $G(q^5)$ and $G(r^2)$. Carefully looking at the Pohlig-Hellman shows that we can actually reduce the DL in $G(q^5)$ to 5 DL's in $\mathbb{Z}_q$. Similarly we reduce the DL in $G(r^2)$ to 2 DL's in $\mathbb{Z}_r$. Hence the remaining problem is solving some DL's in the group $\mathbb{Z}_s$, for $s\in\left\{2,p,q,r\right\}$. That is, a prime order cyclic group.
You may start wondering why I'm going through all of this. The point is that the usefulness of Pohlig-Hellman is precisely this reduction step. Assuming that $p,q,r$ are all small, we can easily solve all DL's in our groups $\mathbb{Z}_s$. However, what if $p,q,r$ are not small? Then we have to solve the DL's in $\mathbb{Z}_s$ some different way, but Pohlig-Hellman does not come with any restrictions. The most straightforward way would perhaps be Baby-Step Giant-Step, but we are free to choose any other method.
Therefore in relation to your actual question, any sieving technique that applies to solving a DL in a prime order group, will also apply here. Although I guess it is most commonly known for factoring, there are definitely applications of the NFS to these groups, see e.g. https://cr.yp.to/bib/2003/joux.pdf.
