Proving there do not exist natural numbers $m$ and $n$ such that $7m^2=n^2$ I'm stuck on my proof of this concept and I could use some help on understanding what to do next.

Prove there do not exist natural numbers $m$ and $n$ such that $7m^2=n^2$

Proof will be by contradiction. So, assume there does exist natural numbers such that $7m^2=n^2$ is true.
Let $m=p_1^{r_1} p_2^{r_2 }…p_k^{r_k}$ and $n=p_1^{t_1} p_2^{t_2}…p_s^{t_s}$
Then $m^2=(p_1^{r_1} p_2^{r_2 }…p_k^{2r_k})^2=p_1^{2r_1} p_2^{2r_2 }…p_k^{2r_k}$ and $n^2=(p_1^{t_1} p_2^{t_2}…p_s^{t_s})^2=p_1^{2t_1} p_2^{2t_2}…p_s^{2t_s}$
Substitute this into $7m^2=n^2$ to obtain,
$7(p_1^{2r_1} p_2^{2r_2 }…p_k^{2r_k})=p_1^{2t_1} p_2^{2t_2}…p_s^{2t_s}$
The idea is that all values of m and n chosen to satisfy $7m^2=n^2$ are natural numbers so they therefore can be expressed as a product of primes.
This is where I'm stuck. How do I go about showing the contradiction?
Thanks for the help
 A: 
Let $m=p_1^{r_1} p_2^{r_2 }…p_k^{r_k}$ and $n=p_1^{t_1} p_2^{t_2}…p_s^{t_s}$

Your indexes here are the same implying that $n$ and $m$ have the same prime factors.
Better to write this as $m=p_{a_1}^{r_1} p_{a_2}^{r_2 }…p_{a_k}^{r_k}$ and $n=p_{b_1}^{t_1} p_{b_2}^{t_2 }…p_{b_s}^{t_s}$.

Substitute this into $7m^2=n^2$ to obtain, $7(p_1^{2r_1} p_2^{2r_2 }…p_k^{2r_k})=p_1^{2t_1} p_2^{2t_2}…p_s^{2t_s}$

Okay that means that there is a $7$ on the LHS so there must be a $7$ on the RHS.  So  one of the $p_{b_i}$ must be equal to $7$.  Without loss of generality let's assume that $p_{b_1} = 7$ then we have

$7m^2 = 7p_{a_1}^{2r_1} p_{2a_2}^{2r_2 }…p_{a_k}^{2r_k}=n^2 = 7^{2t_1} p_{2b_2}^{2t_2 }…p_{b_s}^{2t_s}$

Now $2t_1 \ne 1$ and $2t_1 > 1$ so that $7$ on the LHS can't be the only $7$ in the factorization of $7m^2$.  So one of the $p_{a_1}$ must be $7$.  Without loss of generality let's assume that $p_{b_1} = 7$.   then we have
$7m^2 = 7\cdot 7^{2r_1}  p_{2a_2}^{2r_2 }…p_{a_k}^{2r_k}=7^{2r_1 + 1} p_{2a_2}^{2r_2 }…p_{a_k}^{2r_k}=n^2 = 7^{2t_1} p_{2b_2}^{2t_2 }…p_{b_s}^{2t_s}$
But that means $2r_1 + 1 = 2t_1$.  But that's impossible as $2r_1 + 1$ is odd and $2t_1$ is even.
A: The power of $7$ in the decomposition of $7m^2$ in prime numbers is odd, whereas the power of $7$ in the decomposition of $n^2$ in prime numbers is even.
A: Note that if $(m,n)$ solution then $(\lambda n,\lambda m)$ is also solution, so without loss of generality we can suppose $\gcd(n,m)=1$.
Now since $7\mid 7m^2\implies 7\mid n^2$ and since $7$ prime $\implies 7\mid n$
Then $n=7n'$ and $7m^2=(7n')^2\iff m^2=7n'^2$.
But with a similar process then $7\mid m$ which is a contradiction since $\gcd(m,n)=1$.
Therefore no solutions except the trivial one $(0,0)$ exists.
