There was some discussion about the terminology in comments to the question -- I interpret the statement "if $y$ converges to $x$, then $f(y)$ converges to $\infty$" to mean "for all $M\in\mathbb{R}$ there is $\delta > 0$ such that $f(y)>M$ for all $y\neq x$ with $\lvert y-x\rvert < \delta$". Let $S$ be the set of all such points $x$, and call a set that contains uncountably many points of $S$ "uncountably divergent" for short.
I don't think that the $\arctan$ transformation is of any help, since it's just a homeomorphism between $\mathbb{R}$ and $(-\pi,\pi)$ and between $\mathbb{R}\cup\{\infty\}$ and $(-\pi,\pi]$ and doesn't change the substance of the required proof. [Edit, after seeing Nick Kirby's answer and following his links: The transformation does make a difference, in that the limit $\infty$ is not allowed as a function value here, whereas it is if it is finite. This difference is relevant with respect to the example of the function with removable discontinuities at all rationals given here, since this uses the limit at the rationals as the function value at the irrationals, which we can't do in the present case, so that this example doesn't yield an example of a function "converging to $\infty$" at the rationals -- in fact there is no such function; see the remark about nowhere dense sets below.]
Assume that $S$ is uncountable. First, observe that $[n,n+1]$ must then be uncountably divergent for some $n\in\mathbb{N}$. Without loss of generality, assume $n=0$, so $[0,1]$ is uncountably divergent.
Now we divide the interval in half. Either $[0,1/2]$ or $[1/2,1]$ or both are uncountably divergent. If only one of them is, we discard the other half and replace the interval by the uncountably divergent half and continue dividing it.
Assume that after each subdivision only one half is uncountably divergent. These halves form a chain of nested closed intervals whose intersection contains exactly one point. Thus, except possibly for one point, all points of $S$ in $[0,1]$ lie in one of the discarded halves that each contain at most countably many points of $S$. But there are only countably many of these halves, contradicting the fact that there are uncountably many points of $S$ in $[0,1]$. Hence the assumption is false and the subdivision must encounter an interval of which both halves are uncountably divergent.
Once we encounter such an interval, we denote it by $I$ and recursively apply the entire halving procedure to it (including the part where we discard intervals of which only one half is uncountably divergent). We call the two resulting intervals $I_0$ and $I_1$ and continue applying the halving procedure to them, resulting in intervals $I_{00}$, $I_{01}$, $I_{10}$, $I_{11}$, and so on, where each resulting interval has the property that both its halves are uncountably divergent.
We discard any points of $S$ lying in the discarded halves, as well as all points lying on a boundary of one of the intervals. Since there are only countably many such points, each interval that was uncountably divergent remains so. Note that each remaining point of $S$ in $[0,1]$ now lies in a chain of nested intervals $I$, $I_{i_1}$, $I_{i_1i_2}$, ... with lengths converging to $0$, and the intersection of that chain contains exactly that point.
We now begin with the interval $I$, choose some point $s$ of $S$ in it and find $\delta$ such that $f(y) > 1$ for all $y\neq s$ with $y \in (s-\delta,s+\delta)$. Now $s$ lies in a chain of nested closed intervals with lengths converging to $0$, so one of these intervals is entirely contained in $(s-\delta,s+\delta)$. Denote this interval by $I_i$, where $i$ is a string of binary digits, and choose the one of $I_{i0}$ and $I_{i1}$ which does not contain $s$, so that $f>1$ on this entire subinterval. We now apply the same procedure to this interval, choosing a point of $S$ in it but now choosing $\delta$ such that $f(y) > 2$.
Iterating this procedure, we obtain a chain of nested closed intervals whose lengths converge to $0$, and the intersection of these intervals contains exactly one point $t$. But now $f(t)>n$ for all $n\in\mathbb{N}$, which is impossible. Hence $S$ must be countable.
[Edit: Note that uncountability was only used to make sure that we have points in both halves of an interval at every level of subdivision -- thus, we can use the same proof to show that $S$ is nowhere dense (except for the "removable technicality" that we discarded the set of boundary points, i.e. the set of points with finite binary representation, which is itself dense -- this was just to ensure that each point of an interval is contained in exactly one of the closed subintervals, but we can alternatively ensure that by choosing one of the four sub-subintervals, at most two of which contain the given point.). The same is not true for the case of finite limits (i.e. limits which are allowed as function values elsewhere), as the example of the function with removable discontinuities at all rationals shows. The crucial difference is in the last step of the proof, since $f(t)>n$ for all $n\in\mathbb{N}$ is impossible, whereas the corresponding statement for the limit $0$, $\lvert f(t)\rvert<1/n$ for all $n\in\mathbb{N}$, is consistent.]