Is 2018 special because of these properties?

I discovered that: $$2018=(6^2)^2+(5^2)^2+(3^2)^2+(2^2)^2$$ We also have: $$13^2+43^2=2018$$

And we have:

$$2018=44^2+9^2+1^2$$

I somehow tend to believe that there could be a finite number of these numbers that are sum of two squares, three squares and four fourth powers.

So we have a system of three Diophantine equations:

$$n=a^2+b^2$$

and $$n=c^2+d^2+e^2$$

and $$n=f^4+g^4+h^4+i^4$$

where, $n,a,b,c,d,e,f,g,h,i \in \mathbb N$.

Is there a finite number of these numbers?

Edit : Also, it is $$2018=35^2+26^2+8^2+7^2+2^2$$ a sum of five squares.

And of $$2018=11^3+7^3+7^3+1^3$$ four cubes.

There are an infinite number of such integers.

One way to show this is to start from the fact that there are an infinite number of Pythagorean triples $a^2 + b^2 = c^2$. Given any such triple, let $N$ be given by:

$$N = (ab)^4 + (bc)^4 + (ac)^4 + c^4$$

By an identity of Fauquembergue (1) we have (given $a^2 + b^2 = c^2$):

$$(ab)^4 + (bc)^4 + (ac)^4 = (a^4 + a^2b^2 + b^4)^2$$

Hence:

$$N = (a^4 + a^2b^2 + b^4)^2 + (c^2)^2$$

Furthermore:

$$(c^2)^2 = (a^2 + b^2)^2 = a^4 + 2a^2b^2 + b^4 = (a^2 - b^2)^2 + (2ab)^2$$

Hence:

$$N = (a^4 + a^2b^2 + b^4)^2 + (a^2 - b^2)^2 + (2ab)^2$$

Reference:

1) Dickson L E History of the Theory of Numbers Vol 2 Ch XXII p 658

One infinite set is $2018k^4$ for any natural $k$. I strongly suspect that there are plenty more. Numbers that are a sum of three squares are very common. Numbers that are a sum of two squares are not so rare, so I would just start picking sums of four cubes and try to satisfy the other two. Another example is $$1^4+2^4+3^4+6^4=1394=2\cdot 17 \cdot 41 \equiv 4\pmod 8$$ so is a sum of two and three squares.