Perfect numbers are $6, 28, 496, 812, \dots $
From this source
Here “double proportion” means that each number is twice the preceding number, as in $1, 2, 4, 8, ….$ For example, $1 + 2 + 4 = 7$ is prime; therefore, $7 × 4 = 28$ (“the sum multiplied into the last”) is a perfect number.
Now following this rule $3$rd perfect number should be
$1+2+4+8=15$ and $15×8=120$ but $120$ is not next perfect number.
Can someone help me where I am going wrong?
The problem here is that $$1+2+4+8=15$$$15$ is not prime. However, in the upper example, $$1+2+4=7$$$7$ is prime. This is mentioned in your text. See on your second line, the quote
[...] $1+2+4$ is prime; therefore [...]
So $$1+2+\dots+2^{s}=2^{s+1}-1$$ has to be prime in order for your rule to hold. As @b00nheT said in the comments, the sum should continue until it becomes a prime.
You happen to be looking at Mersenne Primes. These can occur when you have $n=2^p-1$, where $p$ is some prime. In your example, $2^3-1=7$ which is prime. However, these never occur when looking at $p$ is not prime. To take your other example, $2^4-1=15$ which is not prime, because $p$ is not prime.
An even number greater than $1$ is perfect iff it is of the form $(2^p-1)2^{p-1}$ where $2^p-1$ is prime. (This can be shown by elementary methods.) So when $p=4,$ we have $2^p-1=15$, which is composite, and $(2^p-1)2^{p-1}=(15)(8)$ is not perfect.
Mersenne primes are primes of the form $2^p-1.$ In order that $2^p-1$ is a prime it is necesary (but not sufficient) that $p$ is prime. For example $2^{11}-1=2047=(23)(89).$ It is unknown whether there is a largest Mersenne prime.
Every perfect number $\mathcal{P}$ is of the form $2^{n - 1}(2^n - 1)$ such that $n$ must be prime and $2^n - 1$ is also prime. Any prime of the form $2^n - 1$ is known as a Mersenne Prime, it's appropriate symbol being $M_n$ (in most cases $M_p$ where $n = p =$ prime).
When you do $1 + 2 + 4 + 8 = 15$, you are proposing that $2^{4 - 1}(2^4 - 1)$ is a perfect number since there are four terms on the $LHS$ (Left Hand Side of the equation). Here I will write it out in a clearer way: $$\sum_{k = 0}^n2^k = x$$ This is read as:
"The sum that goes from $2^0$ to $2^n = x$." Here the variable $k$ is used to express how it starts from $0$ then goes to $1$, $2$, $3$, $\ldots$ $n$. $$\sum_{k = 0}^n2^k = 2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n = x$$ This $\longrightarrow \sum$ is known as a "Capital Sigma" (where the lower case sigma is $σ$.) "Sigma" along with other fancy symbols are mainly derived from the Greek alphabet.
To spare you the math, $$\sum_{k = 1}^n2^k = x = 2^n - 1$$ So when you do $$1 + 2 + 4 + 8 = 2^0 + 2^1 + 2^2 + 2^3 = \sum_{k = 0}^32^k = 15$$ and then multiply the furthest $RHS$ (Right Hand Side of the equation) by $2^3 = 8$, you will not get a perfect number since $15$ is not prime! $15 = 3\times 5$.
A nice trick is to know what $\sum_{k = 0}^n2^k$ is, is to just count how many numbers are there from $0$ to $n$ including $0$ and $n$ themselves. Let's call that number $d$. Then, $$\sum_{k = 0}^n2^k = 2^{d - 1}$$ And since $k$ is every number from $0$ to $n$ then all you need to do is count how many terms there are on the $LHS$.
Example:
$$\sum_{k = 0}^{64}2^k = 2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^{64} = 2^{65} - 1$$ Now there is no way in telling if $2^{65} - 1$ is prime just by looking at it... unless you know that if you want $2^n - 1$ to be prime, then $n$ must also be prime. And since $65$ is obviously divisible by $5$ (and $13$), then $2^{65} - 1$ is not prime, or at least a Mersenne Prime, so $$2^{65 - 1}(2^{65} - 1) = 2^{64}(2^{65} - 1) \neq \mathcal{P}$$ such that $\mathcal{P}$ is a perfect number. I hope now you can find a solution for $n$ in the following equation: $$\sum_{k = 0}^n2^k = \frac{496}{2^{n - 1}}$$ And perhaps by just making $496$ the subject (making it alone on the $RHS$) you can figure out why $\sum_{k = 0}^n2^k = 2^n - 1$ if you haven't already. Hope I helped! :)