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I want to know in which situations two numbers that can be expressed as powers can be equal. I think it's intuitive that if two powers (say $a^b$ and $c^d$) are equal, then the bases must be themselves powers of a single natural number (say $a=m^k$ and $c=m^l$, so that $m^{kb}=m^{ld}$, and $kb=ld$). But I'm having a hard time proving it.

My approach is as follows: I factor $a$ and $c$ into prime factors and write the equation $a^b=c^d$ as:
$$left({p_1}^{a_1} {p_2}^{a_2} cdots {p_n}^{a_n}right)^b=left({p_1}^{c_1} {p_2}^{c_2} cdots {p_n}^{c_n}right)^d$$
(where $p_1=2$, $p_2=3$, $p_3=5$, etc are the prime numbers) which turns into the following system:
$$a_1 b = c_1 d \ a_2 b = c_2 d \ vdots \ a_n b = c_n d$$
I want to prove that the vectors $left[a_1,a_2,ldots,a_nright]$ and
$left[c_1,c_2,ldots,c_nright]$ are both integer multiples of some other vector. It's ok to assume that the mdc's of boths lists are $1$, because if, say, $mathrm{mdc}left{a_1,a_2,ldots,a_nright}=k>1$, then we could factor out $k$ from the $a_j$'s and incorporate it into $b$ (this is the same as reducing $a^b$ to $m^{kb}$). Assuming that the mdc's are $1$, I want to prove that the vectors for $a$ and $c$ are equal.

I don't where to go from here. I can do it in the case that the mdc's are pairwise $1$, that is, $mathrm{mdc}left{a_1,a_2right}=mathrm{mdc}left{a_1,a_3right}=cdots=mathrm{mdc}left{a_{n-1},a_nright}=1$, but not in the general case.

You notice the fact about the gcd's (i.e., mdc's) of the numbers $a_i$ and $c_i$. The key here is to notice that $b$ and $d$ can also be assumed relatively prime (if they have gcd equal to $k$, then take the $k$th root of both sides of $a^b = c^d$ to reduce to the case where the exponents are relatively prime.)

Now the key fact is known as Euler's Lemma: if $b$ and $d$ are relatively prime and $b$ divides evenly into $md$, then $b$ is a divisor of $m$.

Assuming this, we find from $a_i b = c_i d$ that $b$ is a divisor of $c_i d$, so by Euler's Lemma is a divisor of $c_i$ for all $i$. Similarly, $d$ is a divisor of $a_i$ for all $i$. Dividing all of the equations $a_i b = c_i d$ by $b$, we find $a_i = e_i d$ for some integers $e_i$. Substituting this into $a_i b = c_i d$ and dividing by $d$, we find as well that $c_i = e_i b$ for all $i$. Thus, the vectors $[a_1, ldots, a_n]$ and $[b_1, ldots, b_n]$ are both multiples of $[e_1, ldots e_n]$.

Alternatively, we first prove the following claim. To use the claim below, let $k:=gcd(b,d)$. From $a^b=c^d$, we have $$a^{frac{b}k}=c^{frac{d}k},.$$ As $gcdleft(dfrac{b}{k},dfrac{d}{k}right)=1$, by the claim, there exists $einmathbb{Z}_{>0}$ such that $$a=e^{frac{d}k}text{ and }c=e^{frac{b}k},.$$

Claim Let $x,y,p,qinmathbb{Z}_{>0}$ be such that $gcd(p,q)=1$ and $x^p=y^q$. Then, there exists $u in mathbb{Z}_{>0}$ such that $x=u^q$ and $y=u^p$.

Proof. As $gcd(p,q)=1$, there exist $r,sinmathbb{Z}$ such that $pr+qs=1$. Hence, $$x=x^{pr+qs}=x^{pr}x^{qs}=left(x^pright)^rx^{qs}=left(y^qright)^rx^{qs}=left(x^sy^rright)^q,.$$ Take $u:=x^sy^r$, which is a rational number. Then, $x=u^q$ and $y=u^p$. Since $x$ is an integer, $u$ must also be an integer (there are multiple ways to show this, the easiest one I know is that $z=u$ is a rational root of $z^q-x$, and by the Rational Root Theorem, we conclude that $uinmathbb{Z}$).

Let us look case by case

1. Either $b$ divides $d$ ( or $d$ divides $b$, it is similar ) then we get $a_i=frac{d}{b}c_i$. Thus your claim is satisfied as for vector $v=[c_1,c_2..c_n]$ your vector $a=frac{d}{b}v$ and $c=v$.


2. $b$ and $d$ are co-primes then $a_i=k_id$ and $c_i=k_ib$ ( here $k_i$ are integers ). Again your claim is satisfied as for vector $v=[k_1,k_2...k_n]$ your vectors $a=dv$ and $c=bv$.


3. Neither divides the other but their gcd is not 1 ( that is they are not co-primes either ). Lets say gcd($b,d$)=$q$. So $frac{b}{q}$ and $frac{d}{q}$ are co-primes. Now again this is similar to case 2 as $a_i=k_ifrac{d}{q}$ and $c_i=k_ifrac{b}{q}$ ( similar to case 2 we can find the vector $v=[k_1,k_2...k_n]$ and vectors $a=frac{d}{q}v$ and $c=frac{b}{q}v$ ).
   
   As these 3 cases are the only possible cases ( also no two cases can occur at once ) your claim is proved.
   
   Hope it helps.