The official solutions (and the other problems) can be found from the web page of European Mathematical Cup.

Problem EMC 2018 P4 wrote: Let $x,y,m,n$ be integers greater than $1$ such that \[ \underbrace{ x^{x^{{\cdot}^{{\cdot}^x}}} }_{m \ \text{times}} = \underbrace{ y^{y^{{\cdot}^{{\cdot}^y}}} }_{n \ \text{times}} \]Does it follow that $m = n$? Remark: This is a tetration operation, so we can also write $^m x = ^ny$ for the initial condition. Great Exp-NT and Size Problem. The answer is $\boxed{yes}$. Suppose otherwise, that $m \not= n$. Then $x \not= y$. Claim 01. $x = y^r$ for some rational number $r$. Proof. Take a prime number $p \mid x$, then $p \mid y$ (easily proved that $\text{rad}(x) = \text{rad}(y)$, we then have \begin{align*} ^m x &= ^n y \\ ^{m-1} x \cdot \nu_p(x) &= ^{n - 1} y \cdot \nu_p(y) \\ \frac{\nu_p(x)}{\nu_p(y)} &= \frac{ ^{n - 1} y}{ ^{m - 1} x} \end{align*}Thus, we conclude that for any prime $p \mid x$, then $\frac{\nu_p(x)}{\nu_p(y)} = c$, which is a constant. Thus, we conclude that \[r = \prod_{p \mid x,y} p^c = \prod_{p \mid x,y} p^{\frac{\nu_p(x)}{\nu_p(y)}} = x^{\frac{1}{y}} \]for a constant rational number $r$. Take a positive integer $z$ such that $x = z^a$ and $y = z^b$ where $a,b \in \mathbb{N}$ and $\gcd(a,b) = 1$. Claim. Either $a = 1$ or $b = 1$. Proof. Now, WLOG $a > b$. Thus, we have \[ a \cdot z^{a \cdot ^{m - 2} x }= b \cdot z^{b \cdot ^{n - 2} y } \]This forces $\frac{a}{b} = z^{b \cdot ^{n - 2} y - a \cdot ^{m - 2} x} $, which means that $\frac{a}{b} = z^k$ for some $k \in \mathbb{Z}$. Since $a > b$, this means that $k > 0$, which forces $k \in \mathbb{N}$, and hence $\frac{a}{b} = z^k$ for some $k \in \mathbb{N}$, forcing $\frac{a}{b} \in \mathbb{N}$, or $b \mid a$. But $\gcd(a,b) = 1$, and so $b = 1$. Claim. We have $x = y^{y^k}$ for some $k \in \mathbb{N}$. Proof. WLOG $x > y$, then we conclude that $x = y^{\ell}$ for some positive integer $\ell \not= 1$. Now, we have \begin{align*} ^m x &= ^n y \\ \ell \cdot y^{ \ell \cdot ^{m - 2} x} &= y^{ ^{n - 2} y} \\ \ell &= y^{ ^{n - 2} y - \ell \cdot ^{m - 2} x } \end{align*}Hence, $\ell = y^k$ for some integer $k$, and since $\ell,y \in \mathbb{N}$, and $^{n - 2} y - \ell \cdot ^{m - 2} x \in \mathbb{Z}$, then $\ell = y^k$ for some $k \in \mathbb{N}$. Hence, we conclude that $x = y^{y^k}$ for some $k \in \mathbb{N}$. Finishing. To finish this, we notice that \begin{align*} ^m x &= ^n y \\ x^{x^{ ^{m - 2} x}} &= x^{x^k \cdot ^{n - 1} y} \\ x^{ ^{m - 2} x } &= x^k \cdot x^{x^k \cdot ^{n - 2} y} \\ ^{m - 2} x &= k + x^k \cdot ^{n - 2} y \end{align*}for some $k \in \mathbb{N}$, $x,y,m,n \ge 2$. Now, notice that $^{m - 2} x > x^k \cdot ^{n - 2} y > x^k$, which means that $x^k \mid ^{m - 2} x$. This forces \[ x^k \mid ^{m - 2} x - x^k \cdot ^{n - 2} y = k \]But since $x \ge 2$, we know that $x^k \ge 2^k = (1 + 1)^k \ge 1 + k$ for $k \in \mathbb{N}$ by Bernoulli's Inequality. Thus, this can't happen if $m \ge 3$, forcing $m = 2$. But, this gives us a quick contradiction since

(I found this problem after making it and solving it, only to find that it already existed)

CircleInvert wrote:

(I found this problem after making it and solving it, only to find that it already existed) SAME IKR I only proved it for $x^x=y^{y^y}$ has no positive integer solutions other than $(1,1)$

see AMM E 3364