This has been asked many times, and in multiple variations, but here is one link for reference:
Group of Order 105 Having Normal Sylow 5/7 Subgroups
In my attempt, everything is identical up till the point where we show one of the two desired subgroups is normal. From here, my approach differs.
Suppose $n_5 = 1$. Let $H$ be the Sylow $5$-subgroup. Suppose by contradiction that $n_7 = 15$. Let $K_1, K_2, \dots, K_{15}$ be the Sylow $7$-subgroups. Since $H$ is normal, $HK_{i}$ is a subgroup for each $i$. Since the intersection of Sylow $p$-subgroups is trivial, we would have $15$ subgroups of order $5 \cdot 7 = 35$. Clearly this will exceed the value of $105$ giving us a contradiction. This seems to be following the theorem for product of subgroups with one of them normal being a subgroup.
This seems too good to be true, so I tried to think of what could be wrong. The only thing I could think of is that perhaps there is only one subgroup of order $35$. Since all of the Sylow $7$-subgroups are conjugate, perhaps the multiplication yields the same subgroup? I don't believe this is stated in the theorem I'm using, but I could possibly see this happening. After all, it just says that HK is a subgroup, not necessarily a unique subgroup. I would have to put some thought into this, but I could see this happening. Even if this is true though, we would have \begin{align} (3-1) \cdot n_3 &{}+ (5-1) \cdot n_5 + (7 - 1) \cdot n_7 + |HK| \\ &= 1 + 2 \cdot 1 + 4 \cdot 1 + 6 \cdot 15 + 35 \\ &= 132 > \lvert G \rvert \end{align}
This seems like a much easier way to come to the conclusion. Almost too easy.
Note that $n_3 = 1$ or $7$ and clearly the case of $7$ would make the above even larger. The same approach can be used if $n_5 = 21$ and $n_7 = 1$.
Is this approach flawed? And if so, what is the reason it is flawed?
It seems you want to argue that the various $HK_i$ have trivial intersection, but this need not be true. Compare with the situation in $S_3$ with $H$ the unique 3-Slow and the $K_i$ the three 2-Sylows.