Does all proofs of irrationality are the same way: "Find an integer between $0$ and $1$"? Excluding square/cubic etc roots of numbers, I've read proofs about the irrationality of $e,\pi,\ln 2,\zeta(2)$ and $\zeta(3)$. In all of them is: assume $x=p/q$, where $x$ is the number trying to proved irrational and $p,q$ integers.
Long story short, by assuming $x=p/q$ we found an integer between $0$ and $1$.
Is no other way to prove a number is irrational? Maybe finding a combination of $p,q$ that is not integer. For example, assume $x=p/q$ then $p+q = \text{something not integer}$.
PS. The only proof that I've that does not end this way is the beuker's proof for $\pi$, where he find the estimation $1/p^n<1/n!$ which is not possible thus $\pi$ is irrational.

In general for logarithms there is also another way. Let's prove that given $a,b\in \Bbb{N}$ such that $b$ is not a power of $a$ then $\log_a(b)\notin \Bbb{Q}$.
Premise
First let's prove this lemma:

Given $x,n\in \Bbb{N}$, if $x$ is not a $n$-th power, then $x^{\frac 1n}\notin \Bbb{Q} $

If this wasn't true then:
$$x^{\frac 1n}=\frac ND \ \ \ \ \ \gcd(N,D)=1 \ \ \ D\neq 1$$
This would imply that:
$$x=\frac{N^n}{D^n}$$
Since $x\in \Bbb{N}$, then $D|N$, but this is absurd, because they are coprimes for hyphotesis. So the lemma is true
Proof
Let's suppose that the statemente isn't true, so:
$$\log_a(b)=\frac ND \ \ \ \ \ \gcd(N,D)=1 \ \ \ D\neq 1$$
For the definition of logarithm:
$$a^{\frac ND}=b \Rightarrow a=(b^D)^{\frac 1N} $$
Since $a\in \Bbb{Q}$ , $b^D$ must be a $n$-th power. But this implies that $N|D$ and this is absurd, because they are coprimes for hyphotesis. So the initial statement is true.
Improvement
Now note that given a number $j\in \Bbb{R}/\Bbb{N}$  then:
$\log_a(b)=\frac{\log_j(b)}{\log_j(a)}$
Since $\log_a(b)$ is irrational, at least one among $\log_j(b)$ and $\log_j(a)$ must be irrational. These are some ideas, maybe with some other effort you can prove the more general case.
