So in the weak induction, we have to show the inductive step
(1) If $P(n)$ is true then $P(n+1)$ is true.
In "strong induction", the inductive step is a bit more liberal.
(1') If $P(i)$ is true for all $m\leqslant i\leqslant n$ then, $P(n+1)$ is true.
In both inductions you need to then prove the "base case"
$P(m)$ is true.
Usually this $m=0$ or $m=1$
You are free to assume much more in the strong induction.
Strong induction is useful if you want to prove something but the induction step doesn't necessarily follow the $P(n)\implies P(n+1)$ framework such as things about divisibility and statements about multiplicative structures.
An example is when you have the statement
Lemma: Any positive integer $n>1$ is a prime or it can be written as a product of primes.
The proof goes like this. Suppose $P(i)$ is true for all $i\leqslant n$. We want to show that $P(n+1)$ is a product of primes.
Case 1: If it is a prime we are done.
Case 2: If it is not, we can write $n+1=ab$ for some positive integers $a,b$. Now, $a,b<n+1$ and so $a,b\leqslant n$ so by our inductive step $a$ and $b$ are primes or can be written as product of primes. So $n+1$ is also a product of primes as well.
(Check the base case which in this case is $n=2$.)
Notice in this case, the weak induction is utterly useless because the factors of $n+1$ have nothing to do with $n$.
Notes on Equivalence: Strong induction and weak induction are logically equivalent under the usual frameworks of mathematics. It should be clear that
Strong induction $\implies$ Weak Induction
The non-trivial direction is to show the converse. But the gist is like this. Often, people use this analogy of Induction as a Domino. Weak Induction is to say that if you the $n$th block is knocked over, the $n+1$-th block will be knocked over as well. Now, if the $n$th block is knocked over, all blocks before that up to some point have been knocked over. (That is the reason why we can assume much more in strong induction.)