Prove that if $r,s,t \in \Bbb{N}$, then $2^r + 2^s = 2^t \iff r=s$

I'm trying to prove the following statement:

If $\{r,s,t\} \subset\Bbb{N}$ then $$2^r + 2^s = 2^t \iff r=s.$$

(We assume $0 \notin \Bbb{N}$ for this problem)

Here is how I attempted to prove that $r = s \implies 2^r + 2^s = 2^t$:

$r=s \implies 2^r + 2^s = 2^r + 2^r = 2*2^r = 2^{r+1}$ and $r \in \Bbb{N} \implies r+1 \in \Bbb{N}$, so the equality holds for $t = r+1$

(I'm not sure this is correct, so a verification of this would be helpful)

Now I need to prove $2^r+2^s = 2^t \implies r=s$.

This is what I've tried:

$2^r + 2^s = 2^t \implies (2^r+2^s)(2^r-2^s) = 2^t(2^r-2^s)$

$\implies 2^{2r}-2^{2s} = 2^{r+t} - 2^{s+t}$

$\implies 2^{2r} - 2^{r+t} = 2^{2s} - 2^{s+t}$

$\implies 2^r(2^r - 2^t) = 2^s(2^s - 2^t)$

And I'm stuck here. I thought that maybe this could help because I have $2^r$ and $2^s$ on different sides of the equality and they are both multiplied by their difference with $2^t$. Can anyone help me?

I'm supposed to do this without using logarithms, powers that are not in $\Bbb{N}\cup\{0\}$, or calculus.

Assume $r < s$. Then $$2^s < 2^r + 2^s < 2^s + 2^s = 2^{s+1}$$ so $2^r + 2^s$ cannot be a power of $2$.
If $r\neq s$, then (interchanging the roles of $r$ and $s$ if necessary) we can assume that $r<s$. Then $$2^r+2^s=2^r(1+2^{s-r})$$ is divisible by an odd integer $>1$ (namely $1+2^{s-r}$), so cannot be a power of $2$.
Let $r<s$. Thus, $r<t$ and we have $$1+2^{s-r}=2^{t-r},$$ which gives $1$ divisible by $2$, which is contradiction.
By the same way we'll get the contradiction for $r>s$.
Thus, $r=s$ and easy to see that it's possible.