$x^3+y^4=7$ has no integer solutions

I am trying to prove that $x^3+y^4=7$ has no integer solutions, but i have no idea how to start, please helps. I have tried to consider mod 7 to restrict the number of possible $x^3$ because $x^3 \equiv -1,0,1 \pmod{7}$, but it is not working.

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Consider the equation modulo $13$. Then $x^3$ can be $0,1,5,8,12$ and $y^4$ can be $0,1,3,9$. None of these add to $7$ modulo $13$.

I chose $13$ because $3|\phi(13)$ and $4|\phi(13)$, so I could get restrictions on both $x^3$ and $y^4$.

universalset’s answer is right, in these problems a useful heuristic is to find a number that produces remainders such that both sides cannot become equal. Here is a short Python program to find such numbers for this problem.

for m in range(2,100): #We hope to find such number less than 100
    a= set([x**3%m for x in range(0,1000)]) #Hoping remainders will cycle somewhere less that 1000 
    b= set([y**4%m for y in range(0,1000)]) #Hoping remainders will cycle somewhere less that 1000 

    flag = 0;
    for x in a:
        if (7%m+x%m)%m in b:
            flag=1;
            break;

    if flag==0:
        print (m,a,b)