Edit: Swapped the cube root and the square root in the display. Fixed. This also altered the subsequent argument.
To use your method, you never want to raise to a power anything that has both $\sqrt[3]{5}$ and $\sqrt{7}$ in it. So, you isolate these one at a time.
Suppose $\sqrt[3]{5} + \sqrt{7} \in \Bbb{Q}$ so there are $p,q \in \Bbb{Z}$ with $q \neq 0$ such that $\sqrt[3]{5} + \sqrt{7} = p/q$. By reducing $p/q$ to lowest terms, we may assume that $\gcd(p,q) = 1$. Since solving quadratics is harder than solving linears (for the second isolation), we start by cubing. We compute $$\begin{align}
\sqrt[3]{5} &= p/q - \sqrt{7}, \\
5 &= (p/q - \sqrt{7})^3 \\
&= p^3/q^3 - p^2\sqrt{7}/q^2 + 7p/q - 7\sqrt{7}, \text{ and}\\
\sqrt{7} &= \frac{5-p^3/q^3 - 7p/q}{-p^2/q^2 - 7} \\
7 &= \left( \frac{5q^3 - p^3 - 7pq^2}{-p^2q - 7q^3} \right)^2, \text{ so} \\
7q^2 \left( p^2 + 7q^2 \right)^2 &= \left( p^3 + 7pq^2 - 5q^3 \right)^2 .
\end{align}$$
Now work modulo $4$. The squares are congruent to $0$ or $1$, so the left-hand side is either $0$ or $3 \pmod{4}$ and the right-hand side is either $0$ or $1 \pmod{4}$. So, either $q$ or $p^2 + 7q^2$ is even and $p^3 + 7pq^2 - 5q^3$ is even.
If $q$ is even, $p^3 + 7pq^2 - 5q^3 \cong p^3 \pmod{4}$. Since $p^3 + 7pq^2 - 5q^3$ is even, we must have $p$ is even, contradicting our assumption that $p/q$ was reduced to lowest terms.
Otherwise, we have $p^2 + 7q^2$ is even. Note that $p^3 + 7pq^2 - 5q^3 = p(p^2+7q^2)-5q^3$, so is even only if $q$ is even. But then $p^2 + 7q^2$ is even only if $p$ is and again we contradict reduction to lowest terms.