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}
The unfortunate fact is that the result in question cannot be proved constructively.
Here "constructive proof" as usual means "a proof using the methods that are acceptable in mathematical constructivism".
If this result could be prove constructively, then we would have a constructive proof that there is a total function $f(x) = N$ from $\mathbb{R}$ to $\mathbb{N}$, where $N$ is the unique integer with $N \leq x < N+1$. But that function is discontinuous, and any function from the real line that we can form constructively will be continuous.
In fact, any function from $\mathbb{R}$ to $\mathbb{N}$, which we can prove constructively is a function, must be constant. See the section 'The continuum' in this Stanford Encyclopedia entry.
It may be easier to think of the following issue, which is the root cause of the facts above. Suppose that we have a quickly-converging Cauchy sequence of rationals, $(a_n)$, that converges to a real $x$. We want to read off $f(x)$ by looking at the entries of $(a_n)$. Suppose that we see $a_k = 1-2^k$ for all the vales of $k$ we have examined. Then the sequence $(a_n)$ might continue its pattern forever, converging to $1$. Or, perhaps, at some moment, the sequence $(a_n)$ could jump "above" $1$ or jump "below" 1. So the evaluation map that takes a quickly converging Cauchy sequence $(a_n)$ and produces $f(\lim a_n)$ will not be continuous. That is the real obstacle to a constructive proof of this result, assuming the reals are defined as Cauchy sequences.
Best Answer
Suppose that there's at least two integers such that the property holds, and suppose that $M<N$.
Now, $M<N \le x <M+1<N+1$ according to the property, but we have that $M<N<M+1$ , but clearly because they're integers (is there an integer $N$ such that there is integers $M$ and $M+1$ and $M<N<M+1$ holds) , $M=N$ should also hold and we have a contradiction.