For every real $x>0$ and every integer $n>0$, there is one and only one real $y>0$ such that $y^n=x$
What you are asking for is the existence of $n$-th root. This is true in $\mathbb{R}$ but not in $\mathbb{Q}$ (for example $\not \exists q\in \mathbb{Q}:q^2=2$). Any proof of this will require the completeness of the real line (Least Upper Bound Property).
LUB property: Every subset of $\mathbb{R}$ bounded above must have a supremum.
Here is a typical proof of this theorem only using the LUB property and the Binomial Theorem.
Let $x>1$ and $S=\left\{ a>0: a^n<x \right\}$. Obviously $1\in S$ and so $S\neq \emptyset $.
Since $x>1\Rightarrow x^n>x$ we have that $S$ is bounded above by $x$. Therefore, by the Least Upper Bound Property, $\exists \sup S=r\in \mathbb{R}$.
We shall prove that $r^n=x$.
Suppose that $r^n>x$ and let \begin{equation}\epsilon =\dfrac{1}{2}\min \left\{ 1,\frac{r^n-x}{\sum\limits_{k=1}^n{\binom{n}{k}r^{n-k}}} \right\}\end{equation} Then $0<\epsilon <1$ and so $\epsilon ^n<1$.
Therefore, by the Binomial Theorem, \begin{gather}(r-\epsilon )^n=\sum\limits_{k=0}^{n}{\dbinom{n}{k}r^{n-k}(-\epsilon )^k}=r^n-\epsilon \sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}\epsilon ^{k-1}(-1)^{k-1}\ge \\ r^n-\epsilon \sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}\epsilon ^{k-1}> r^n-\epsilon \sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}\\ (r-\epsilon)^n> r^n-\dfrac{r^n-x}{\displaystyle\sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}} \displaystyle\sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}=r^n-r^n+x=x\Rightarrow \left( r-\epsilon \right)^n>x\end{gather} which is a contradiction since $r=\sup S$ and $r-\epsilon <r$
Suppose that $r^n<x$ and let \begin{equation}\epsilon =\dfrac{1}{2}\min \left\{ 1,\frac{x-r^n}{\sum\limits_{k=1}^n{\binom{n}{k}r^{n-k}}} \right\}\end{equation} Then, $0<\epsilon <1$ and so $\epsilon ^n<1$. Therefore, by the Binomial Theorem, \begin{gather}(r+\epsilon )^n=\sum\limits_{k=0}^{n}{\dbinom{n}{k}r^{n-k}\epsilon^k}=r^n+\epsilon \sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}\epsilon ^{k-1}<\\ r^n+\epsilon \sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}\\ (r+\epsilon)^n< r^n+\dfrac{x-r^n}{\displaystyle\sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}} \displaystyle\sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}= r^n+x-r^n=x\Rightarrow (r+\epsilon)^n<x\end{gather} and so $\sup S=r<r+\epsilon\in S$ which is a contradiction.
Therefore, $r^n=x$ if $x>1$
Let $0<x<1$. Then $\frac{1}{x}>1\Rightarrow \exists r>0:r^n=\frac{1}{x}\Rightarrow \exists r'=\frac{1}{r}>0:r'^n=\frac{1}{r}^n=\frac{1}{r^n}=x$.
If $x=1$ then $r=1$.
EDIT: Motivation as per request: As I said this is statement is not true if we replace $\mathbb{R}$ with $\mathbb{Q}$. What sets $\mathbb{R}$ and $\mathbb{Q}$ apart is the completeness of $\mathbb{R}$. The LUB property therefore must in some way be used. This is why we define $S$. Because the LUB is an existensial theorem, it shows the existence of a supremum but not its value, it is often used in proofs by contradiction.
So assuming $r^n>x$ we need to arrive to a contradiction. Remember $r$ is a very special number, the supremum of $S$. If we could show that $\exists m\in \mathbb{R}$ so that $m<r$ and is an upper bound of $S$ or $m>r$ and $m\in S$ then we are done. This is what we do with $m=r-\epsilon$ in the first case and with $m=r+\epsilon$ in the second.
It all boils down to finding an $\epsilon>0$ so that $r-\epsilon$ is an upper bound of $S$, that is $(r-\epsilon)^n>x$. We can make this $\epsilon$ as small as we want. I shall choose an $\epsilon$ for $n=2$. \begin{equation}(r-\epsilon )^2=r^2-2r\epsilon+\epsilon^2>r^2-2r\epsilon-\epsilon\end{equation} Remember we want $(r-\epsilon )^2>x$ and so it suffices \begin{equation}r^2-2r\epsilon-\epsilon>x\Leftrightarrow \epsilon<\frac{r^2-x}{1+2r}\end{equation}
Assume that $y^n=z^n$ with $y,z>0$, $n\in \mathbb N$, $n>1$. Then $$0=y^n-z^n=(y-z)(y^{n-1}+y^{n-2}z+y^{n-3}z^2+\ldots + yz^{n-2}+z^{n-1}).$$ The secod parentheses is a sum of $n-1$ positive summands, hence nonzero. Therefore the other factor $y-z$ must be zero, i.e. $y=z$ and thus there can be at most one $n$th root of $x>0$.
For existence, $f\colon[0,x+1]\to\mathbb R$, $y\mapsto y^n$ is continuous and we have $f(0)=0<x$ and $f(x+1)=(x+1)^n=x^n+nx^{n-1}+\cdots +n x + 1>nx\ge x$, hence by IMV, there exists an $y$ with $f(y)=x$. You might in fact even try a very direct approach and show that $\{q\in\mathbb \mid q>0, q^n>x\}$ defines a Dedekind cut ...