I would say you have to use two things :
1- That the Hurewicz morphism is a morphism of long exact sequences for a pair $(X,A)$ of spaces (apply this to, e.g. $(M_{\tilde f},\tilde X)$, where $M_{\tilde f}$ is a mapping cylinder for $\tilde f$)
2- That the Hurewicz theorem is slightly more precise than "if all homotopy groups $\leq n-1$ vanish, then it's an iso in degree $n$" : it also says that in this case, the Hurewicz morphism is surjective in degree $n+1$.
So then you have the following commutative diagram (where I assume, wlog thanks to the cylinder, that $\tilde f : \tilde X\to \tilde Y$ is the inclusion of a subspace):
$\require{AMScd}\begin{CD}\pi_{n+1}(\tilde Y,\tilde X)@>>> \pi_n(\tilde X)@>>> \pi_n(\tilde Y) @>>> \pi_n(\tilde Y,\tilde X)\\
@VVV @VVV@VVV@VVV \\
H_{n+1}(\tilde Y,\tilde X) @>>> \tilde H_n(\tilde X)@>>> \tilde H_n(\tilde Y)@>>> H_n(\tilde Y,\tilde X)\end{CD}$
I have hidden the rest of the details so that you can try to write it down yourself.
You know that the two outermost maps in the top row are $0$ because $\tilde f$ is an iso on $\pi_n$, that the rightmost vertical map is an iso, that $\pi_n(\tilde Y,\tilde X)=0$ and that the leftmost vertical map is an epimorphism. This is enough to conclude.
${}$
Indeed it follows that $H_n(\tilde Y,\tilde X) = 0$; and then let $x\in H_{n+1}(\tilde Y,\tilde X)$, it comes from $y\in \pi_{n+1}(\tilde Y,\tilde X)$ by the epimorphism, which is sent to $0$ in $\pi_n(\tilde X)$, and then to $0$ in $\tilde H_n(\tilde X)$, so by commutativity, $x$ is also sent to $0$. Therefore $H_{n+1}(\tilde Y,\tilde X)\to \tilde H_n(\tilde X)$ is $0$, and $\tilde H_n(\tilde Y)\to H_n(\tilde Y,\tilde X)$ too, so this allows us to conclude.
Best Answer
Relative hurewicz applied to $\tilde f$ shows that it is an isomorphism on homotopy groups, but $\pi_n \tilde X = \pi_n X$ for $n>1$, and this isomorphism commutes with the maps induced by $f$ and $\tilde f$.
So your map $f$ is an iso on all homotopy groups. Apply Whitehead.