Lemma: $f(k)>k$.
Pf: it is clear that $f(1)≥1$ and that $f(1)\neq 1$ so the claim is true for $k=1$. Inductively suppose it is true up to $k-1$. Then $f(k-1)>k-1$ so $f(k-1)≥k$. Since $f(k)>f(k-1)$ we are done.
Claim: $f(n)=n+1$
Pf: Indeed the Lemma shows that $f(n)≥n+1$. But $f(f(n))=n+2\implies f(n)≤n+1$ and we are done.
The only completely multiplicative increasing functions $\mathbb{N} \to \mathbb{N}$ are of the form $f(n) = n^j$ for some positive integer $j$.
To prove this, let $f(2) = 2^w$ for some real number $w$. Then consider $f(k)$.
Suppose that $f(k) > k^w$. In this case, we have $\log_2(f(k)) > w \log_2(k)$ and therefore $\log_2(k) < \frac{\log_2(f(k))}{w}$.
The rational numbers are dense, so take some $a \in \mathbb{N}, b \in \mathbb{N}, b > 0$ such that $\log_2(k) < \frac{a}{b} < \frac{\log_2(f(k))}{w}$.
Then we see that $k^b < 2^a$. Therefore, $f(k)^b = f(k^b) < f(2^a) = f(2)^a = 2^{w \cdot a}$. But we also see that $2^{w \cdot a} < f(k)^b$. Contradiction.
Similarly, suppose $f(k) < k^w$. In that case, we have $\log_2(k) > \frac{\log_2(f(k))}{w}$.
By the density of the rational numbers, we take some $a, b \in \mathbb{N}$, $b > 0$ such that $\frac{\log_2(f(k))}{w} < \frac{a}{b} < \log_2(k)$.
From $\frac{a}{b} < \log_2(k)$, we derive that $2^a < k^b$ and hence that $2^{a \cdot w} = f(2)^a = f(2^a) < f(k^b) = f(k)^b$. From $\frac{\log_2(f(k))}{w} < \frac{a}{b}$, we derive that $f(k)^b < 2^{a \cdot w}$. Again, we have a contradiction.
Therefore, we see that $f(k) = k^w$ for all $k$.
Clearly, we must have $w > 0$ to make $f$ increasing. And since $n^w$ is an integer for all $n$, we see that $w$ must be an integer.
So the answer is that the only possible values of $f(2)$ are the even powers of 2.
Best Answer
You have already shown the cases $n=1,2$. Now we want to do the induction step. To simplify notation I write $m=n+1$. If $m$ is not prime, we are done (as we can just use the induction hypothesis on the factors). Thus, in the following we will assume that $m$ is prime. First we show a lower bound
$$ f(m)^2=f(m^2) > f(m^2-1)=f(m+1) f(m-1) = (m+1)^2 (m-1)^2 = (m^2-1)^2.$$ We used that $m$ is prime and thus $m\pm 1$ are divisible by $2$, which implies that we can use the induction hypothesis to show $f(m\pm 1)=(m\pm 1)^2$. Taking square root yields $f(m)>m^2-1$.
Next we show the upper bound. Let $N\in \mathbb{N}_{\geq 1}$, then there exists a unique $\ell(N)\in \mathbb{N}$ such that $$ 2^{\ell(N)} \leq m^N < 2^{\ell(N)+1}.$$ Then we have $$ m^N < 2^{\ell(N)+1} \leq 2 m^N. $$ Thus, we obtain $$ f(m)^N < f(2^{\ell(N)+1}) = (2^{\ell(N)+1})^2 < 4 m^{2N}. $$ After taking the $N$th root, we get $$ f(m) \leq 4^{1/N} m^2. $$ Hence, for $N\rightarrow \infty$ we get $$ f(m) \leq m^2. $$ Combining the upper and the lower bound we get $f(m)=m^2$.