Show that if for prime $p$ there exists $m,n \in \mathbb{N}$ such that $p^2=2^n 3^m + 1$, then $p\leq 17$ 
Show that if for prime $p$ there exists $m,n \in \mathbb{N}$ such that $p^2=2^n 3^m + 1$, then $p\leq 17$.

I have no idea even how to begin approaching this problem.

Suppose $p$ is a prime with the property that the only prime factors of $p-1$ and $p+1$ are $2$ and $3$
We assume $p>17$
Then, one of the numbers $p-1$ and $p+1$ contain only the prime factor $2$, hence is a power of $2$.
Case $1$ : $p-1$ is a power of $2$. Then $p=2^n+1$ and $p+1=2^n+2=2\cdot (2^{n-1}+1)$. 
Case $2$ : $p+1$ is a power of $2$. Then $p=2^n-1$ and $p-1=2^n-2=2\cdot (2^{n-1}-1)$
We have $n>4$ because of $p>17$
One of the numbers $2^{n-1}+1$ and $2^{n-1}-1$ has no prime factor other than $3$, and because of $n>4$ . we have $2^{n-1}-1\ge 15>9$. 
So, one of the numbers $2^{n-1}+1$ and $2^{n-1}-1$ must be a non-trivial power of $3$. Hence, we get a solution of the equation $2^m-3^n=\pm1$. 
Since it is well known that the only solution with $m,n>1$ is $m=3$ and $n=2$ (Catalan's conjecture, now proven) , we get a contradiction.
