How to solve this system with equation and inequality? { <mtable columnalign="right

Well, as suspected, your edit brings a new piece of information (the inequation) that allows to easily solve that system using only high-school mathematics (the role of the inequality being to rule out the "ugly" non-zero solutions) - with the exception of a single step in the proof that seems to require very high-level mathematics.
For simplicity, let $t=2^x$. The equation implies that $x^2={\displaystyle \frac{16}{49}}(t-<!--\; -\; -->1)$. Plugging this into the inequality and performing all the necessary simplifications brings it to
$(t-<!--\; -\; -->1)(49-<!--\; -\; -->56t+16t^2)\le <!--\; \le \; -->0.$
Inside the second pair of brackets one recognizes $(7-<!--\; -\; -->4t{)}^2$, so there are only two possibilities:
- either $t-<!--\; -\; -->1\ge <!--\; \ge \; -->0$ and $(7-<!--\; -\; -->4t{)}^2\le <!--\; \le \; -->0$
- or $t-<!--\; -\; -->1\le <!--\; \le \; -->0$ and $(7-<!--\; -\; -->4t{)}^2\ge <!--\; \ge \; -->0$
Since $(7-<!--\; -\; -->4t{)}^2\ge <!--\; \ge \; -->0$, the first possibility implies that $(7-<!--\; -\; -->4t{)}^2=0$ so that $t={\displaystyle \frac{7}{4}}$, which also satisfies $t-<!--\; -\; -->1\ge <!--\; \ge \; -->0$. Returning to $x$, this has the consequence that $2^x={\displaystyle \frac{7}{4}}$ and $x^2={\displaystyle \frac{12}{49}}$, or equally well $x={\log }_2<!--\; \; -->7-<!--\; -\; -->2$ and $x=\pm <!--\; \pm \; -->{\displaystyle \frac{2\sqrt{3}}{7}}$. Since $x={\log }_2<!--\; \; -->7-<!--\; -\; -->2\ge <!--\; \ge \; -->{\log }_2<!--\; \; -->4-<!--\; -\; -->2=0$, it follows that x cannot take that negative value. Can it take the positive one? Assuming it could, this would mean that ${\log }_2<!--\; \; -->7=2+{\displaystyle \frac{2\sqrt{3}}{7}}$, which after exponentiation would be equivalent to ${\displaystyle \frac{7}{4}}=4^{{\displaystyle \frac{\sqrt{3}}{7}}}$. This, in turn, implies that $4^{{\displaystyle \frac{\sqrt{3}}{7}}}\in <!--\; \in \; -->\mathbb{Q}$, which after raising to the $7$th power implies that $4^{\sqrt{3}}\in <!--\; \in \; -->\mathbb{Q}$. This is not true, but I do not know how to prove this at a high-school level. Maybe you are supposed to use a scientific calculator to solve your problem? Anyway, that $4^{\sqrt{3}}$ is irrational is a consequence of the Gelfond-Schneider theorem, or of the Lindemann-Weierstrass theorem, both of them, though, being vastly out of the reach of high-school pupils. In any case, the conclusion so far is that $t={\displaystyle \frac{7}{4}}$ does not lead to solutions of the given system.
The second possibility reduces to $t-<!--\; -\; -->1\le <!--\; \le \; -->0$, because the square is always $\ge <!--\; \ge \; -->0$, but since $x^2={\displaystyle \frac{16}{49}}(t-<!--\; -\; -->1)$, this would imply that $x^2=0$ ($0$ being the only number both $\ge <!--\; \ge \; -->0$ and $\le <!--\; \le \; -->0$), which has the only solution $x=0$.
The analysis could have been performed equally well by replacing $t$ with ${\displaystyle \frac{49}{16}}{x}^2+1$ and performing all the compuations with $x$, instead of with $t$, but the reasoning would have reached the same difficult point about ${\log }_2<!--\; \; -->7$ that has been reached above.