The equation is
$u^3(s^4 + (r-1)^4) - s^4(t - 1)^3 = 0$
Has no integer solutions for $u,s \neq 0$. How do mathematicians today approach this problem?
Sorry for broadness, just looking for a general direction. Thanks.
With the variable substitution $q = r -1 , \ v = t-1$, we have that the above equation has a solution in $u,s,r,t \iff u^3s^4 + u^3q^4 - s^4v^3 = 0$ has a solution in $u,s,q,v$. There's a bijective correspondence between the solution sets.
This comes from the Beal conjecture: $x^a + y^b - z^c = 0$ has no coprime integer solutions $x, y, z \gt 0$ for odd prime $a,b,c$.
It has a counter example iff there exist $r,s,t,u \in \Bbb{Z}$ such that $x^a - y^b - z^c = 0$ and $rx + sy = 1$ and $tx + uz = 1$ and $x, y, z \gt 0$, $a,b,c$ odd primes. Proof. $\Rightarrow$. If any two of $x,y,z$ share a gcd $\neq 1$, then they all share it so are not coprime, therefore $\gcd(x,y) = 1$ and $\gcd(x,z) = 1$ and the result follows. $\Leftarrow$. If such $r,s,t,u$ exist then $x, y, z$ are coprime and solutions to the equation so, we're done.
So we have Beal counter-example $x,y,z, a,b,c$ iff there exists $r,s,t,u \in \Bbb{Z}$ such that :
$$ (1) \\ x^a + y^b - z^c = 0 \\ rx + sy = 1 \\ tx + uz = 1 $$
Notice that $r,s,t,u$ can't be zero, so we can divide by them. So we have the above equations imply $$ x^a + (\frac{1 - rx}{s})^b - (\frac{1 - tx}{u})^c = 0 $$
has a solution for $1 - rx, s, 1 - tx, u, x \gt 0$. And the converse holds. With the variable substitution $q = 1 - rx$, $v = 1 - tx$ we have that there's a Beal counterexample iff $u^c s^b x^a + u^c q^b - s^b v^c = 0$ for $u,s,x,q,v \gt 0$.
Thus Beal is equivalent to the conjecture that $u^c s^b x^a + u^c q^b - s^b v^c = 0$ has no integer solutions $u,s,x, q, v \gt 0$, $a,b,c$ odd primes, with $q = 1 \pmod {r}$, and $v = 1 \pmod {t}$.
Note that this "answer" does not settle the (nontrivial) part of your equation, only shows it is really the same as either of two other simpler ones.
I'll consider your equivalent equation $$u^3(s^4+q^4)=s^4v^3.\tag{1}$$ To begin with one should disregard the case $q=0$ as in that case it becomes $u^3=v^3$ which holds when $u=v$. Note first that one may assume $\gcd(u,v)=1$ since otherwise any common factor can be removed from both sides of $(1).$ For the same reason one may assume $\gcd(s,q)=1,$ and note that also implies $\gcd(s^4,s^4+q^4)=1.$ Now from these coprime assumptions it follows that $u^3 | s^4$ and also $s^4 |u^3,$ from which $u^3=s^4$ and your equation simplifies to $$s^4+q^4=v^3. \tag{2}$$ I don't have a proof that this has no nonzero solutions. But it looks similar to the Beale conjecture except for the power $4$ not being prime. So perhaps the algebra that led to $(1)$ didn't really change the problem much. [I wonder: Did you simply start with $(2)$ and manipulate it into $(1)$ by use of Bezout expressions of gcd conditions? If so as this shows the problem did not really change by that algebra.]
Note also that since we had $u^3=s^4$ we can rewrite $(2)$ as $$q^4=v^3-u^3.\tag{3}$$ To sum up, your equation is equivalent to asking if the sum of two fourth powers can be a cube, and also to asking whether a single fourth power can be a difference of cubes. I freely admit both of these questions are beyond my ability.