There are several loose yet elementary bounds on the prime numbers. Say for an example, an (very loose indeed...) upper bound on $p_n$, the $n^{th}$ prime number: $$p_n<2^{2^{n}}$$ can be easily achieved by induction (appendix 1). A much more tight one, $$p_n<4^n$$, could also be proved by using elementary combinatorics (appendix 2).
So, this post is to collect upper or lower bounds on $p_n$ which could also be attained by elementary approaches (by the way, there is a urban legend saying that the prime number theorem has an elementary proof, but not yet verified for me), which is (in an educator's view, maybe) indeed a very good way to inspire students who are studying elementary number theory to some beautiful method from an eye-catching result.
Thanks in advance.
Appendix 1
Induction: the proposition $p_n<2^{2^n}$ is obviously true for $n=1$. Now, assuming that it is true for some $k\in\mathbb N$, we have
$$p_{k+1}\le 1+p_1p_2\ldots p_k< 1+2^{2^1+2^2+\ldots +2^k}$$ $$=1+2^{2^{k+1}-2}<2^{2^{k+1}}$$ Which completed the proof with the principle of mathematical induction.
Appendix 2
Let $q$ be the $n^{th}$ prime number ($q=p_n$). Now, it's easy to show that every positive integer $k\le q$ could be written in the form $$k=m^2 p_1^{e_1}p_2^{e_2}\ldots p_n^{e_n}$$ Where $m<\sqrt q$ might be a composite number or a prime, and $e_i$ is either $1$ or $0$. So, since the strict upper bound of the number of naturals we could express by the above formula is $\sqrt q 2^n$, it gives $$q<\sqrt q2^n\implies q< 4^n$$ As desired.
Let's begin with a variation of Erdős' argument that yields a tighter bound. Since every squarefree number $\leqslant x$ is the product of some set of distinct primes $\leqslant x$, we have
$$2^{\pi(x)} \geqslant Q(x).$$
Erdős' argument effectively uses a lower bound of $\sqrt{x}$, but we can easily show $Q(x) \geqslant \frac{x}{2}$, since
$$Q(x) \geqslant \lfloor x\rfloor - \sum_{p} \biggl\lfloor\frac{x}{p^2}\biggr\rfloor > x\Biggl(1 - \sum_p \frac{1}{p^2}\Biggr) - 1,$$
where the numbers divisible by the square of more than one prime are subtracted more than once (so the first inequality is strict for $x \geqslant 36$, compensating for the final $-1$). It's not hard to see $\sum_p \frac{1}{p^2} < \frac{1}{2}$, for
\begin{align} \sum_{p} \frac{1}{p^2} &= \frac{1}{4} + \frac{1}{9} + \frac{1}{25} + \sum_{p \geqslant 7} \frac{1}{p^2} \\ &< \frac{361}{900} + \sum_{k = 3}^{\infty} \frac{1}{(2k+1)^2} \\ &< \frac{361}{900} + \frac{1}{2}\sum_{k=3}^{\infty}\biggl(\frac{1}{2k(2k+1)} + \frac{1}{(2k+1)(2k+2)}\biggr) \\ &= \frac{361}{900} + \frac{1}{12} \\ &= \frac{109}{225}, \end{align}
and then one can verify $Q(x) \geqslant \frac{x}{2}$ for $x < 36$ by hand. Thus we have $2^{\pi(x)}\geqslant \frac{x}{2}$, which yields $\pi(x) \geqslant \frac{\log x}{\log 2} - 1$, and taking $x = p_n$ it yields $p_n \leqslant 2^{n+1}$. A more precise analysis of $Q(x)$ can only reduce the constant $\pm 1$, since we obviously have the upper bound $Q(x) \leqslant x$.
Unfortunately I don't know any elementary argument that shows $\pi(x) \geqslant x^{\alpha}$ for some $\alpha \in (0,1)$ [and doesn't yield the correct $x/\log x$ behaviour], so the next step is already giving the correct order of magnitude.
Chebyshev had the brilliant idea of using the central binomial coefficients to estimate $\pi(x)$. This works very well because they have a simple prime factorisation that we know exactly (for a suitable value of "know exactly"), and we know their magnitude.
Using $\lfloor 2y\rfloor - 2\lfloor y\rfloor \in \{0,1\}$ for all $y\in \mathbb{R}$ and Legendre's formula for the exponent of $p$ in the prime factorisation of factorials, we see that the exponent of $p\leqslant 2n$ in the factorisation of $\binom{2n}{n}$ is
$$e_p := \sum_{k : p^k \leqslant 2n} \Biggl(\biggl\lfloor \frac{2n}{p^k}\biggr\rfloor - 2\biggl\lfloor \frac{n}{p^k}\biggr\rfloor\Biggr) \leqslant \frac{\log (2n)}{\log p},$$
and hence
$$\binom{2n}{n} = \prod_{p \leqslant 2n} p^{e_p} \leqslant \prod_{p\leqslant 2n} p^{\log (2n)/\log p} = (2n)^{\pi(2n)}.$$
Using $\binom{2n}{n} \geqslant \frac{2^{2n}}{2n}$ this yields
$$\pi(2n) \geqslant \frac{2n\log 2}{\log (2n)} - 1,$$
and choosing $n = \bigl\lceil \frac{x}{2}\bigr\rceil$ we obtain
$$\pi(x) \geqslant \log 2\frac{x}{\log x} - 2$$
for $x \geqslant 2$. (Credit goes to fedja, whose excellent answer drove home the point that the proof is actually simple.)
To get an upper bound for $\pi(x)$, we note that every prime $n < p \leqslant 2n$ divides $\binom{2n}{n}$ with exponent $1$, whence
$$\prod_{n < p \leqslant 2n} p \leqslant \binom{2n}{n} \leqslant 2^{2n}.$$
Thus
$$\vartheta(2^k) - \vartheta(2^{k-1}) = \log \prod_{2^{k-1} < p \leqslant 2^k} p \leqslant 2^k\log 2,$$
and summing these estimates yields
$$\vartheta(2^k) < 2^{k+1}\log 2.$$
Choosing $k$ so that $x \leqslant 2^k < 2x$, we have
$$\vartheta(x) \leqslant \vartheta(2^k) < 2^{k+1}\log 2 \leqslant (4\log 2) x.$$
Now a summation by parts yields
\begin{align} \pi(x) &= \sum_{p \leqslant x} (\log p)\cdot \frac{1}{\log p} \\ &= \frac{\vartheta(x)}{\log x} + \int_2^x \frac{\vartheta(t)}{t(\log t)^2}\,dt \\ &\leqslant 4\log 2\Biggl(\frac{x}{\log x} + \int_2^x \frac{dt}{(\log t)^2}\Biggr) \\ &= 8 + 4\log 2\operatorname{Li}(x). \end{align}
With slightly more careful estimates, one can reduce the constant factor to $2\log 2$, and with some serious work one can get much closer to $1$, both for the upper and lower bounds.