If we multiply through by $pab$, we find that for $a, b \ne 0$, our equation is equivalent to $pb+pa=ab$, which we can rewrite as
$$(a-p)(b-p)=p^2.$$
We are looking for non-zero solutions of this equation. Conveniently, $p^2$ does not have many factorizations!
We can have $a-p=-1$, $b-p=-p^2$, which yields a negative $b$, or $a-p=-p^2$, $b-p=-1$, which yields a negative $a$.
We could have $a-p=-p$, $b-p=-p$, but that yields the impossible $a=b=0$.
We can have $a-p=1$, $b-p=p^2$, or $a-p=p^2$, $b-p=1$, which respectively yield the solutions $a=p+1$, $b=p^2+p$, and $a=p^2+p$, $b=p+1$.
Finally, we can have $a-p=p$, $b-p=p$, which yields $a=b=2p$.
Comment: The same idea can be used to find all integer solutions of the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$, where $n$ is a given positive integer. The factorizations of $n^2$, with the exception of $n^2=(-n)(-n)$, supply the integer solutions.
Because $p,q,r,s,t$ are consecutive positive integers, we know immediately that $q=r-1,p=r-2,s=r+1,t=r+2$. This means that $q+r+s=r-1+r+r+1=3r$ and that $p+q+r+s+t=5r$.
If $3r$ is a perfect square, then this means that the prime factorization of $r$ must have $3^{2n+1}$ for some $n$; similarly, for $5r$ to be a perfect cube, we must have that the prime factorization of $r$ must have $5^{3m+2}$ for some $m$.
Clearly having other factors will lead to $r$ being larger, so we can say that $r$ is of the form $r=3^{2n+1}5^{3m+2}$. But $3r$ being a perfect square means that $3m+2$ must be even, so $m$ is even; the least such $m$ is 0. Similarly, $5r$ being a perfect cube means that $2n+1$ must be a (positive) multiple of 3; the least such $n$ to make this possible is $n=1$.
It follows that $r=3^35^2$.
Best Answer
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.