Find all the natural numbers $n$ such that $\sigma(n)=15$ 
Find all the natural numbers $n$ such that $\sigma(n)=15$ 
Where $\sigma (n)$ is the sum-of-divisors function

My attempt:
$$n=p_1^{\alpha_1}\cdots p_s^{\alpha_s}$$
$$\sigma(n)=\frac{p_1^{\alpha_1+1}-1}{p_1-1}\cdots \frac{p_s^{\alpha_s+1}-1}{p_1-1}=15$$
$1.\quad \text{ for }s=1:$
$$n=p^\alpha,\quad \frac{p^{\alpha+1}-1}{p-1}=15$$
$$p^{\alpha+1}-1=15p-15\\
p^{\alpha+1}=15p-14$$
I don't know what I should do now.

Here is a way to finish the s = 1 case in the way you started it and continuing from your last line:
$(p^\alpha - 15)p = 14$.
Since p is prime and divides 14 we must have either $p = \pm 2$ or $p =  \pm 7$. So we are left with four cases: $2^\alpha - 15 = 7$, $2^\alpha - 15 = -7$, $7^\alpha  - 15 = -2$ and $7^\alpha  - 15 = 2$. It is easy to see that in only one case $\alpha$ is an integer, which means that there is exactly one solution to the $s = 1$ case.
Time to move on to s = 2. Here it useful that there are not that many ways to write 15 as the product of 2 integers. So we can either solve the system of two equations:
$(p^{\alpha + 1} - 1)/(p-1) = 15$ and $(q^{\alpha + 1} - 1)/(q-1) = 1$
or the system of two equations
$(p^{\alpha + 1} - 1)/(p-1) = 3$ and $(q^{\alpha + 1} - 1)/(q-1) = 5$
Of course all of them take some work, and after solving this we have to move on two s = 3, but as Daniel points out above, we won't be busy for ever. You will find that in order to solve the s = 3 case you are looking at equations you already studied in the s = 2 case and so on. (Note for instance that of the four equations needed in the s = 2 case, the first has already been studied when working on s = 1.)
Good luck!
(With many thanks to Daniel for below comment!)
