Let the polynomial be $P(x)$. We are given that there exists three distinct integers $a,b,c$ such that $P(a)=1$, $P(b)=2$ and $P(c)=3$. Now, for the sake of contradiction, suppose there exists two distinct integers $m$ and $n$ such that $P(m)=P(n)=5$. Let $Q(x)=P(x)-5$. Then this means, $Q(x)$ has roots $m$ and $n$. Without loss of generality, suppose that $m>n$. By Factor Theorem, there must exist a polynomial $R(x)$ with integer coefficients such that $$Q(x)=(x-m)(x-n)R(x)$$We also know that $Q(a)=-4$, $Q(b)=-3$ and $Q(c)=-2$. This means, $$\begin{cases} -4=(a-m)(a-n)R(a)\\ -3=(b-m)(b-n)R(b)\\ -2=(c-m)(c-n)R(c)\end{cases}$$From the second equation, we know that $(b-m)$ and $(b-n)$ must both be odd. This means, $m$ and $n$ must have the same parity. From the third equation, we know that $(c-m)$ and $(c-n)$ cannot both be even. (Otherwise, the LHS would be divisible by 4). Therefore, $(c-m)$ and $(c-n)$ are both odd integers. This can only happen if $$\begin{cases}c-m=-1\\ c-n=1\end{cases}\implies m-n=2$$Now, we look at the first equation. We have: $$-4=(a-m)(a-n)R(a)$$The only divisors of $-4$ are $-4,-2,-1,1,2,4$. The only ones with a difference of two are $(-4,-2)$, $(-1,1)$ and $(2,4)$. If $\begin{cases}a-m=-4\\ a-n=-2\end{cases}$ or $\begin{cases}a-m=2\\ a-n=4\end{cases}$, then this means $8$ would divide $-4$, which is absurd. If $\begin{cases}a-m=-1\\ a-n=1\end{cases}$, then this means $a=b$, which is also absurd. We see that both cases lead to a contradiction. Therefore, there does not exist distinct integers $m$ and $n$ such that $P(m)=P(n)=5.\blacksquare$