Let $m = 2^a \cdot 3^b \cdot 5^c \cdot 7^d$. $N$ where $N$ is a product of primes different from $2$, $3$, $5$ and $7$. For a number $n$ to be an exact 5th power, all the exponents in the prime factorisation of $n$ must be divisible by $5$. Similarly for 6th and 7th powers.
For primes other than $2$, $3$, $5$ and $7$ the exponent in the prime decomposition of $m$ must be $0$ or a multiple of $5 \times 6 \times 7 = 210$. The smallest solution will occur when each exponent is $0$. Thus we can consider only numbers of the form $2^a \cdot 3^b \cdot 5^c \cdot 7^d$.
For the conditions to be satisfied we must have $5 \mid c+1$, $6 \mid c$ and $7 \mid c$. The smallest solution of this is $c = 84$. To obtain $a$ we need to solve
$$a \equiv 0 \pmod{5} \quad a+1 \equiv 0 \pmod{6} \quad a \equiv 0 \pmod{7}.$$
The smallest solution is $35$. As $b$ satisfies the same equations, the smallest value of $b$ is also $35$. Finally, $d$ has to satisfy
$$d \equiv 0 \pmod{5} \quad d \equiv 0 \pmod{6} \quad d+1 \equiv 0 \pmod{7}.$$
The smallest solution is $d = 90$. Putting everything together, the smallest value of $m$ is $5^{84} \cdot 6^{35} \cdot 7^{90}$.