Suppose there are $m,n\in \mathbb{N}$ such that $\varphi(n)=2\cdot 7^{5m+4}$.
Then $\varphi(n)=\varphi(p_1^{k_1}p_2^{k_2}...p_r^{k_r})=\varphi(p_1^{k_1})\varphi(p_2^{k_2})...\varphi(p_r^{k_r})=p_1^{k_1-1}(p_1-1)p_2^{k_2-1}(p_2-1)...p_r^{k_r-1}(p_r-1)=2\cdot 7^{5m+4}$
In another words, $2=\frac{p_1^{k_1-1}(p_1-1)p_2^{k_2-1}(p_2-1)...p_r^{k_r-1}(p_r-1)}{7^{5m+4}}$. This means that one of $p_i$'s is 2 with $k_i=2$. WLOG, assume that $p_1=2$ and $k_1=2$. Then $p_2^{k_2-1}(p_2-1)...p_r^{k_r-1}(p_r-1)=7^{5m+4}$ but this is not possible since we must have that $p_i=7$ and $p_i-1=6$ also appears.
Does this make sense?
I would suggest the following approach :
If $p$ is a prime factor of $n$, then $p-1$ divides $\varphi(n)=2\cdot 7^{5m+4}$ The only divisors $d$ of $2\cdot 7^{5m+4}$ for which $d+1$ is prime, are $1$ and $2$, since $2\cdot 7^k+1$ is divisible by $3$ and greater than $3$ for every positive integer $k$, whereas $7^k+1$ is obviously not prime for positive integer $k$. Hence $n$ cannot have a prime factor exceeding $3$ and hence its totient value cannot be divisible by $7$.