For how many positive integers $m$ the number $m^3+5m^2+3$ is a perfect cube?

We only need to check a finite number of cases: when $|m|$ becomes large enough $f(m)=5m^2+3$ will always lie between $g(m)=3m^2+3m+1=(m+1)^3-m^3$ and $h(m)=6m^2+12m+8=(m+2)^3-m^3$, and so $m^3+5m^2+3$ is between two consecutive cubes.

$h(m)-f(m)$ is positive for $m\le-12$ and $m\ge0$, while $f(m)-g(m)$ is always positive. Therefore, we only need check $-11\le m\le-1$ inclusive, and this reveals that no $m$ is such that $m^3+5m^2+3$ is a perfect cube.

So we have for some $n$: $$m^3+5m^2+3=n^3$$ Say $n$ and $m$ differ for $k$ then $n=m+k$ so we have $$5m^2+3 = 3m^2k+3mk^2+k^3$$ and now we have a quadratic equation in $m$ with an integer parameter $k$: $$\boxed{m^2(3k-5)+3mk^2+(k^3-3)=0}$$

Now the discirminant is a perfect square so $$9k^4-4(3k-5)(k^3-3)=d^2$$ for some integer $d$. Now check when this polynomial $$p(k)=-3k^4+20k^3+36k-60$$ is positive (and a perfect square) and you are (almost) done.

Since (I used here Am-Gm inequality) $$20k^3+36k\geq 3k^4+60 >2k^4+\underbrace{k^4+3+9+3}_{\geq 36|k|}$$

we have $$k^3(10-k)>0\implies k\in \{1,2,3...,9\}$$

Assume, for a contradiction,that it's a perfect cube and notice : $m^3 < m^3 +5m^2+3 < (m+2)^3$ so we must have $m^3+5m^2+3=(m+1)^3$ but that equation has no integer solutions.