<s>[INST] Solving $4^m-3^n=p^2$ for natural $(m,n,p)$. I know that such exponential equations, like the one in question, $4^m-3^n=p^2$ (where $m,n,p$ are natural) are usually solved using numerical methods, so I tried the following.
My attempt
I’ve split the search into parts; at least intuitively (and WolframAlpha confirms), the only solutions are $(0,0,0)$ and $(1,1,1)$. So, I’ve shown that these are the solutions for $m,n\le 2$ by direct calculation. Then, I analysed the last digits of the square difference in each case, first noting that:
$$l(4^k)=\begin{cases}
4, 2\mid k\\
6, 2\nmid k
\end{cases}$$
And that:
$$l(3^k)=\begin{cases}
3, k\equiv 1\: (\text{mod }4)\\
9, k\equiv 2\: (\text{mod }4)\\
7, k\equiv 3\: (\text{mod }4)\\
1, 4\mid k
\end{cases}$$
Where $l$ is the last-digit function. For $2\mid m$ and $n\equiv 1 \text{ or } 2\: (\text{mod }4)$ and for $2\nmid m$ and $k \equiv 3 \text{ or } 0\: (\text{mod }4)$, the last digits are $3$ or $7$, so $p$ cannot be a natural number.
However, I am not sure how I should go about the other $4$ cases. Any hints on how to continue with this proof, or does anybody know a simpler / nicer approach?
 [/INST] Hint: Write like this 
$$(2^m-p)(2^m+p) =3^n$$

Solution:
So $2^m-p = 3^a$ and $2^m+p = 3^b$ for some $a,b$ where $a+b=n$ and $b\geq a\geq 0$. 
Now 
$$ 2^{m+1}= 3^b+3^a = 3^a(3^{b-a}+1)\implies a=0  \;\;\;{\rm and}\;\;\; 2^{m+1}= 3^b+1$$
so 
$$(-1)^{m+1}\equiv_3 1 \implies  m+1=2k$$
so $$ (2^k-1)(2^k+1)=3^b$$ and we can repeat story $2^k-1 =3^x$ and $2^k+1=3^y$ for some $x,y$ where $x+y=k$ and $y\geq x\geq 0$.
So $$2 =3^y-3^x = 3^x(3^{y-x}-1)\implies x=0 \;\;\;{\rm and}\;\;\;3^y=3\implies y=1$$
So $k=1$ and $m=1$. So $n=1$ and $p= 1$. 

If $0$ is also natural number then for $b=0$ we have $n=0$ and $m=0$ and $p=0$.  
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