Claim: $\log_2(2015)$ is irrational
Proof: Let's assume it is rational. Then:
$$\log_2(2015)=a/b, \quad a,b\in \mathbb Z \setminus \{0\}, b\neq 0$$
Question 1: is it okay how I chose a,b?
then we get: $2^{a/b}=2015 \Leftrightarrow 2^a=2015^b$
Now we know: $2015=31\cdot13\cdot5$ so we get:
$$2^a=(31\cdot13\cdot 5)^b=13^b \cdot 13^b \cdot 5^b \tag{$\ast$}$$
Since we know that any integer can be represented uniquely by a multiplication of primes, we would have two ways of representing the same number in such a way. So it wouldn't be unique. Thus, it isn't rational. Thus, it is indeed irrational.
Question: I don't like the step ($\ast$). I see that the argument at the end is indeed what we want but I'm not sure if it really holds, since we have one exponent for several different primes. E.g. one could also get: $2^p=2^q\cdot 31 \cdot 13 \cdot 5$ Which makes it absolutely clear.
Question: Does ($\ast$) hold?
What you did is correct (other than at a passage where you wrote $13$ instead of $31$). But it is simpler to say that $2^a$ is even and $2015^b$ is odd.