Is this a valid argument to show that for $m > 4$, $2^m - 3^n \ne 7$

Question

Is the follow argument valid? Is there a simpler argument?

Here's the argument:

(1) Assume that there exists for integer $m > 4, n$: $$2^m - 3^n = 7$$

(2) $m$ and $n$ are even since:

$$2^{m} - 3^{n} \equiv 4 - 9 \equiv 7 \pmod {12}$$

(3) So, there exists integers $i,j$ with $m=2i, n=2j$ with:

$$2^m - 3^n = (2^i - 3^j)(2^i + 3^j) = 7$$

(4) But, since $i > 2$, $2^i - 3^j \ne 1$ from the Catalan's Conjecture

(5) Then, both $2^i - 3^j$ and $2^i + 3^j$ are greater than $1$ which is impossible since $7$ is prime. So we reject our assumption in step(1).

Answer 1

Catalan’s conjecture is way overkill.

Since $2^i-3^j<2^i+3^j$ are integer and the second is positive, the only way for their product to be $7$ is if $2^i+3^j=7.$ But $m\geq 6$ means $i\geq 3,$ so $$2^i+3^j>2^3=8>7.$$

However, Catalan might be needed to prove the more general result that there is no prime $p\equiv 7\pmod{12}$ and $m>4,n>0 $ such that:

$$p=2^m-3^n$$

Edit: (From commenter Erick Wong) It turns out, it is pretty easy to show there is no $2^i-3^j=1$ with $i>2$ just by looking modulo $8,$ so you don’t need Catalan for that, either.

Answer 2

Just a small alternative to Thomas Andrews's answer: If $(2^i-3^j)(2^i+3^j)=7$ then $2^i-3^j=1$ and $2^i+3^j=7$, which implies $2\cdot2^i=1+7=8$, or $i=2$ (corresponding to $m=2i=4$).