Can different tetrations have the same value? Suppose, we have two numbers
$a\uparrow \uparrow b$ and $c\uparrow \uparrow d$. 
To avoid trivial cases, suppose $a,b,c,d>2$ and $(a,b)\ne (c,d)$.

Is there a quartupel $(a,b,c,d)$ with $a\uparrow\uparrow b=c\uparrow \uparrow d$ under the given conditions ?

In general, $b<d$ implies $a\uparrow \uparrow b<c\uparrow\uparrow d$, but if $a$
is large enough, $a\uparrow\uparrow b$ will exceed $c\uparrow \uparrow d$. So, this argument does not rule out the possibility of the equality.
 A: No, that is not possible, not even if we allow one or more of $a,b,c,d$ to be $2$.
Suppose it was possible, assume without loss of generality that $a>c$, and consider the prime factorizations
$$ a = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k} \\
c = p_1^{\gamma_1} p_2^{\gamma_2} \cdots p_k^{\gamma_k} $$
Because $a^{a\uparrow\uparrow (b-1)}=c^{c \uparrow\uparrow(d-1)}$ we must have that $\frac{\alpha_i}{\gamma_i}=\frac{c\uparrow\uparrow(d-1)}{a\uparrow\uparrow (b-1)} =: Q$ for every $i$.
The net exponent of $p_i$ in $\frac{c\uparrow\uparrow(d-1)}{a\uparrow\uparrow (b-1)}$ is $$\gamma_i(c\uparrow\uparrow(d-2))-\alpha_i(a\uparrow\uparrow(b-2))=\gamma_i\Bigl((c\uparrow\uparrow(d-2))-Q(a\uparrow\uparrow(b-2)\Bigr)$$
which has the same sign for every $i$, and since $a>c$, $Q$ must be an integer and $a=c^Q$.
But then $Q=\frac{c\uparrow\uparrow(d-1)}{a\uparrow\uparrow(b-1)}$ becomes
$$ Q = c^{c\uparrow\uparrow(d-2)-Q(a\uparrow\uparrow(b-2))}$$
Setting $R=\log_c Q$, this is
$$ R = c\uparrow\uparrow(d-2)-c^R(a\uparrow\uparrow(b-2)) $$
Now, $c\uparrow\uparrow(d-2)$ and $a\uparrow\uparrow(b-2)$ are both non-negative powers of $c$, so we have
$$ R = c^{\text{something}} - c^{R+\text{something}} $$
which must be positive (otherwise $Q=1$ and $a=c$) and is therefore a multiple of $c^{R+\text{something}}$. But for $R$ to be a multiple of $c^R$ when $c\ge 2$ is a contradiction!
