Find $n \in N$ where $133^5 + 110^5 + 84^5 + 27^5 = n^5$

Question

This question was posited to me by a coworker. I was able to get the answer, but I'm unsatisfied with how I did it.

1. $n < 354$

$n^5 = 133^5 + 110^5 + 84^5 + 27^5 < (133+110+84+27)^5$

$n < 354$

2. $n \equiv 0 \pmod 2$

133 and 27 are odd, 110 and 84 are even. Thus $133^5 + 110^5 + 84^5 + 27^5$ is the sum of two even numbers and two odd numbers, which results in an even number.

3. $n \equiv 0 \pmod 3$

$133 \equiv 1 \pmod 3 \implies 133^5 \equiv 1$

$110 \equiv 2 \implies 110^5 \equiv 2^5 \equiv 32 \equiv 2$

$84 \equiv 0$

$27 \equiv 0$

$133^5 + 110^5 + 84^5 + 27^5 \equiv 1 + 2 + 0 + 0 \equiv 0$

3. $n \equiv 4 \pmod 5$

$133 \equiv 3 \pmod 5$

$110 \equiv 0$

$84 \equiv 4 $

$27 \equiv -3$

$133^5 + 110^5 + 84^5 + 27^5 \equiv 3^5 + 0 + 4 - 3^5 \equiv 4$

$n^5 \equiv 4 \implies n \equiv 4$

4. $n \equiv 24 \pmod {30}$

From Chinese Remainder Theorem:

$n \equiv 0 \pmod 2, n \equiv 0 \pmod 3 \implies n \equiv 0 \pmod 6$

$n \equiv 0 \pmod 6, n \equiv 4 \pmod 5$

$n \equiv 4 * 1 * 6 \pmod {30} \implies n \equiv 24 \pmod {30}$

5. $n \equiv 4 \pmod 7$

$133 \equiv 0 \pmod 7$

$110 \equiv 5 \to 5^2 \equiv 4 \to 5^4 \equiv 2 \to 110^5 \equiv 5^4 * 5 \equiv 3$

$84 \equiv 0$

$27 \equiv -1$

$133^5 + 110^5 + 84^5 + 27^5 \equiv 0 + 3 + 0 - 1 \equiv 2$

$n^5 \equiv 2 \implies n \equiv 4$

6. $n \equiv 144 \pmod {210}$

$n \equiv 24 \pmod {30}, n \equiv 4 \pmod 7$

$n \equiv 4*4*30 + 24*13*7 \pmod {210} \equiv 60 + 84 \equiv 144$

7. The next number after 144 that satisfies $n \equiv 144 \pmod {210}$ is $n = 354$. From 1. we know that $n < 354$. Therefore $n = 144$

What I've proven is that if there's a solution with $n \in N, n = 144$.

I'm left wondering - How do I prove the solution actually exists?