If $f$ is an entire function that goes to infinity, $f$ is a polynomial

Let $g(z)=f\left(\frac1z\right)$. Then $\lim_{z\to0}g(z)=\infty$ and therefore $g$ has a pole at $0$, because otherwise $g$ would have an essential singularity at $0$, which is impossible, by the Casorati-Weierstrass theorem. But if $f(z)=a_0+a_1z+a_2z^2+\cdots$, then $g(z)=a_0+\frac{a_1}z+\frac{a_2}{z^2}+\cdots$ and therefore asserting that $g$ has a pole at $0$ means that $a_n=0$ if $n$ is large enough. So, $f$ is a polynomial function.

I believe you have mis-read or mis-interpreted the key part of the problem. For complex variables, $$ \lim_{z\to \infty} f(z) = X $$ means that the limit is $X$ (whatever $X$ is) as $z$ approaches infinity in any direction.

I'm not certain, but I believe you could replace that with $$ \forall u \in \Bbb C : (|u|=1 \implies \lim_{r\to\infty_{r\in\Bbb R}} f(r u) = X) $$ that is, the limits, taken as the limit of the real length along a line from the origin, of the function of points on that line, are all the same value $X$ (which in this case is $\infty$.

So you need to show that for any non-polynomial entire function, there is some direction along with the function does not go to infinity.