if $3^m-2^m=65$, then $m=$ what?

Here,
$3^m-2^m=65$

Or, $\left(3^m\right)^\frac22-\left(2^m\right)^\frac22=65$

Or, $\left(3\right)^\frac{2m}2-\left(2\right)^\frac{2m}2=65$

Or, $\left(3\right)^{\frac m2.2}-\left(2\right)^{\frac m2.2}=65$

Or, $\left(3^\frac m2\right)^2-\left(2^\frac m2\right)^2=65$

Let, $\left(3^\frac m2\right)=x$

and $\left(2^\frac m2\right)=y$

Now, the equation will be,

$x^2-y^2=65$

Or, $\left(x+y\right)\left(x-y\right)=13\times5$

So, $\left(x+y\right)=13$ ---------Eq($i$)

And, $\left(x-y\right)=5$ ---------Eq($ii$)

Adding Eq($i$) & ($ii$)

$\left(x+y\right)+\left(x-y\right)=13+5$

Or, $x+y+x-y=18$

Or, $2x=18$

Or, $x=\frac{18}2$

$\therefore x=9$

Now, Subtracting Eq($i$) & ($ii$)

$\left(x+y\right)-\left(x-y\right)=13-5$

Or, $x+y-x+y=8$

Or, $2y=8$

Or, $y=\frac82$

$\therefore y=4$

Re-call,

$\left(3^\frac m2\right)=x$

Or, $\left(3^\frac m2\right)=9$

Or, $\left(3^\frac m2\right)=3^2$

Or, $\frac m2=2$

$\therefore m=4$

Try Again, $\left(2^\frac m2\right)=y$

Or, $\left(2^\frac m2\right)=4$

Or, $\left(2^\frac m2\right)=2^2$

Or, $\frac m2=2$

$\therefore m=4$

Proof of solution,

Replacing $m=4$ to the given equation

Or, $3^m-2^m=65$

Or, $3^4-2^4=65$

Or, $3\cdot3\cdot3\cdot3-2\cdot2\cdot2\cdot2=65$

Or, $81-16=65$

$\therefore 65=65$