Yes, you have to put at least $m+1$ pigeons into one hole.
Proof is by contradiction:
Suppose not, so every hole contains at most $m$ pigeons. In that case there are at most $mn < mn+1$ pigeons across the $n$ holes, contradiction.
But we can find a distribution such that no hole contains $m+2$ pigeons:
Put $m+1$ pigeons in the first hole, then $m$ pigeons in the subsequent $(n-1)$-holes. Then there are $m+1+(n-1)m = nm+1$ pigeons in total (as required), but no hole contains $m+2$ pigeons.
Actually, your initial reasoning is a perfectly good instance of 'reasoning by pigeonholing': there are at most 31 'even' holes for pigeons to go in, so with 32 pigeons you're bound to get an odd number.
That's it!
Your second method is far more complicated than it has to be. Yes, you can make it work by making the holes $\{ 0,1 \}$, $\{2,3 \}$, etc. but also by using $\{0,3 \}$, $\{ 1,2 \}$, etc. In fact, to get as many holes as even numbers, you could even use $\{ 0,37,39 \}$, $\{2, 13 \}$, $\{ 4 \}$, $\{ 6, 19,23,29,59 \}$, etc. In other words, adding the odd numbers to the even numbers when all that really counts is how many even numbers there are is completely extraneous.
Now: I understand you tried to set it up in such a way that you can try to answer both the question about the odd and the even numbers at once ... which seemed to work fine ... until you got to the $60$ 'by-itself-hole' ... and now you get into trouble: Using $\{ 60 \}$ as a hole means it can contain exactly one 'even' pigeon, but now it cannot contain any 'odd' pigeon, and so it can't be used to do the reasoning regarding odd numbers, and using $\{ 58,59,60 \}$ means two 'even' pigeons can go into that hole, and so now it cannot be used to reason about the even numbers.
So really the best thing is to answer the question about the even numbers separately from the question about the odd numbers: with $31$ even numbers you need to pick $32$ pigeons to get an odd pigeon, and with $30$ odd numbers you need $31$ pigeons to get an even pigeon.
Best Answer
Although you didn't say so in the question, your title indicates that you want a formula in propositional logic. So I'd express "each pigeon is in a hole" as $$ \bigwedge_{i=1}^{n+1}\bigvee_{j=1}^n p[i,j], $$ and I'd express "some hole contains two distinct pigeons" as $$ \bigvee_{j=1}^n\bigvee_{1\leq i<k\leq n+1}(p[i,j]\land p[k,j]). $$ Finally, I'd express the pigeonhole principle as the implication from the first of these formulas to the second. HOWEVER, some people understand the pigeonhole principle as including the additional hypothesis that each pigeon is in only one hole, which is formulated as $$ \bigwedge_{i=1}^{n+1}\bigwedge_{1\leq j<k\leq n}\neg(p[i,j]\land p[i,k]). $$ As for the length of the formula, that depends not only on whether you include the extra hypothesis but also on the details of the definition of length --- do you count occurrences of atomic formulas, or occurrences of connectives, or both, or something else?