This is true (for large $m$ and $n$) under ABC plus the assumption that there is a prime in $[x,x+x^{1/2-\delta}]$ for some positive $\delta$ (which is widely believed, but beyond RH). To see this, suppose $m <n$ and that they have the same radical $r$. Write $m=gM$ and $n=gN$ where $g$ is the gcd of $m$ and $n$, so that $M$ and $N$ are coprime. Applying the ABC conjecture to $M + (N-M) = N$, we conclude that $$ (N-M) r \ge N^{1-\epsilon}, $$ so that $$ n-m \ge n^{1-\epsilon}/r. $$ On the other hand, clearly $n-m$ is also $\ge r$ (since it is divisible by $r$). It follows that $n-m \ge n^{1/2-\epsilon}$, and the assumption that there are primes in short intervals finishes the (conditional) proof.

The problem is likely very hard, as fedja's observation in the comments already shows. There is a conjecture of Hall that $|x^3-y^2| \gg x^{1/2-\epsilon}$ which is wide open. The best results that are known here (going back to Baker's method) are of the flavor $|x^3-y^2| \gg (\log x)^C$ for some $C$. If $x^3-y^2$ does get as small as in the Baker result, then take $n=x^4y$ and $m=xy^3$, which clearly have the same radical and then $|n-m|$ is of size essentially $n^{5/11}$. In other words, either you have to improve work towards Hall's conjecture, or work towards gaps between primes!

This is true (for large $m$ and $n$) under ABC plus the assumption that there is a prime in $[x,x+x^{1/2-\delta}]$ for some positive $\delta$ (which is widely believed, but beyond RH). To see this, suppose $m <n$ and that they have the same radical $r$. Write $m=gM$ and $n=gN$ where $g$ is the gcd of $m$ and $n$, so that $M$ and $N$ are coprime. Applying the ABC conjecture to $M + (N-M) = N$, we conclude that $$ (N-M) r \ge N^{1-\epsilon}, $$ so that $$ n-m \ge n^{1-\epsilon}/r. $$ On the other hand, clearly $n-m$ is also $\ge r$ (since it is divisible by $r$). It follows that $n-m \ge n^{1/2-\epsilon}$, and the assumption that there are primes in short intervals finishes the (conditional) proof.

The problem is likely very hard, as fedja's observation in the comments already shows. There is a conjecture of Hall that $|x^3-y^2| \gg x^{1/2-\epsilon}$ which is wide open. The best results that are known here (going back to Baker's method) are of the flavor $|x^3-y^2| \gg (\log x)^C$ for some $C$. If $x^3-y^2$ does get as small as in the Baker result, then take $n=x^4y$ and $m=xy^3$, which clearly have the same radical and then $|n-m|$ is of size essentially $n^{5/11}$. In other words, either you have to improve work towards Hall's conjecture, or work towards gaps between primes!

Added Thanks to Pasten's comment, I learned that this problem is already in the literature and is known as Dressler's conjecture. The conditional proof above is recorded in work of Cochrane and Dressler who give more information on the conjecture.

This is true (for large $m$ and $n$) under ABC plus the assumption that there is a prime in $[x,x+x^{1/2-\delta}]$ for some positive $\delta$ (which is widely believed, but beyond RH). To see this, suppose $m <n$ and that they have the same radical $r$. Write $m=gM$ and $n=gN$ where $g$ is the gcd of $m$ and $n$, so that $M$ and $N$ are coprime. Applying the ABC conjecture to $M + (N-M) = N$, we conclude that $$ (N-M) r \ge N^{1-\epsilon}, $$ so that $$ n-m \ge n^{1-\epsilon}/r. $$ On the other hand, clearly $n-m$ is also $\ge r$ (since it is divisible by $r$). It follows that $(n-m) \ge n^{1/2-\epsilon}$, and the assumption that there are primes in short intervals finishes the (conditional) proof.

The problem is likely very hard, as fedja's observation in the comments already shows. There is a conjecture of Hall that $|x^3-y^2| \gg x^{1/2-\epsilon}$ which is wide open. The best results that are known here (going back to Baker's method) are of the flavor $|x^3-y^2| \gg (\log x)^C$ for some $C$. If $x^3-y^2$ does get as small as in the Baker result, then take $n=x^4y$ and $m=xy^3$, which clearly have the same radical and then $|n-m|$ is of size essentially $n^{5/11}$. In other words, either you have to improve work towards Hall's conjecture, or work towards gaps between primes!

This is true (for large $m$ and $n$) under ABC plus the assumption that there is a prime in $[x,x+x^{1/2-\delta}]$ for some positive $\delta$ (which is widely believed, but beyond RH). To see this, suppose $m <n$ and that they have the same radical $r$. Write $m=gM$ and $n=gN$ where $g$ is the gcd of $m$ and $n$, so that $M$ and $N$ are coprime. Applying the ABC conjecture to $M + (N-M) = N$, we conclude that $$ (N-M) r \ge N^{1-\epsilon}, $$ so that $$ n-m \ge n^{1-\epsilon}/r. $$ On the other hand, clearly $n-m$ is also $\ge r$ (since it is divisible by $r$). It follows that $n-m \ge n^{1/2-\epsilon}$, and the assumption that there are primes in short intervals finishes the (conditional) proof.

The problem is likely very hard, as fedja's observation in the comments already shows. There is a conjecture of Hall that $|x^3-y^2| \gg x^{1/2-\epsilon}$ which is wide open. The best results that are known here (going back to Baker's method) are of the flavor $|x^3-y^2| \gg (\log x)^C$ for some $C$. If $x^3-y^2$ does get as small as in the Baker result, then take $n=x^4y$ and $m=xy^3$, which clearly have the same radical and then $|n-m|$ is of size essentially $n^{5/11}$. In other words, either you have to improve work towards Hall's conjecture, or work towards gaps between primes!